{"id":7073,"date":"2024-12-27T18:52:07","date_gmt":"2024-12-27T18:52:07","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=7073"},"modified":"2024-12-27T18:52:07","modified_gmt":"2024-12-27T18:52:07","slug":"faradays-law-of-electrolysis","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/12\/27\/faradays-law-of-electrolysis\/","title":{"rendered":"Faraday\u2019s Law of Electrolysis"},"content":{"rendered":"\n

The relationship between the quantity of electric charge passed through an electrolyte, and the amount of substance deposited at the electrodes was given by Faraday in 1834, in the form of the law of electrolysis.<\/p>\n\n\n\n

Faraday\u2019s First Law<\/strong><\/p>\n\n\n\n

When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to the quantity of electric charge passed through the electrolyte.<\/p>\n\n\n\n

If W is the mass of the substance deposited by passing Q coulomb of charge, then according to this law:<\/p>\n\n\n\n

W \u221d Q<\/p>\n\n\n\n

Now, Q = I \u2715 t<\/p>\n\n\n\n

W \u221d I \u2715 t<\/p>\n\n\n\n

W = z \u2715 I \u2715 t<\/p>\n\n\n\n

Where Z is a constant, known as electrochemical equivalent and is characteristic of a substance deposited.<\/p>\n\n\n\n

Faraday\u2019s constant (F) \u2013 <\/strong>It is the charge possessed by 1 mole of electrons, and it is equal to 96500 coulombs (approx.). In terms of Faraday\u2019s constant, the number of gram equivalent of electrolyte discharged at an electrode is equal to Faraday\u2019s passed,<\/p>\n\n\n\n

W = E X Q \/ 96500<\/p>\n\n\n\n

Faraday\u2019s Second Law<\/strong><\/p>\n\n\n\n

When the same quantity of charge is passed through different electrolytes, then the mass of different substances deposited at the respective electrodes will be in the ratio of their equivalent masses.<\/p>\n\n\n\n

Mathematically, it is represented as,<\/p>\n\n\n\n

W1<\/sub> \/ W2<\/sub> = Z1<\/sub> \/ Z2<\/sub><\/p>\n\n\n\n

Where, W1 <\/sub>and W2 <\/sub>are the weight of two substances which are deposited at their respective electrodes, and Z1<\/sub> and Z2<\/sub> are their respective equivalent weight.<\/p>\n\n\n\n

Solved Problems<\/h2>\n\n\n\n

1. Given that,<\/strong><\/p>\n\n\n\n

Fe2+<\/sup>\/ Fe = -0.44 v, Ni+2<\/sup>\/ Ni = 0.25 v<\/strong><\/p>\n\n\n\n

Ag+ <\/sup>\/ Ag = 0.80 v, Cu+2<\/sup>\/ Cu = 0.34 v<\/strong><\/p>\n\n\n\n

Which of the following reactions under standard conditions will not take place in the specified direction?<\/strong><\/p>\n\n\n\n

(a) Ni+2<\/sup>(aq) + Cu(s) \u2192 Ni(s) + Cu+2<\/sup>(aq)<\/strong><\/p>\n\n\n\n

(b) Cu(s) + 2 Ag+<\/sup>(aq) \u2192 Cu+2<\/sup>(aq) + 2 Ag(s)<\/strong><\/p>\n\n\n\n

(c) Cu(s) + 2H+<\/sup>(aq)\u2192 Cu+2<\/sup> + H2 <\/sub>(g)<\/strong><\/p>\n\n\n\n

(d) Fe(s) + 2H+<\/sup>(aq)\u2192 Fe+2<\/sup> (aq) + 3 H2<\/sub><\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Options a and c.<\/strong><\/p>\n\n\n\n

Ni+2<\/sup>(aq) + Cu(s) \u2192 Ni(s) + Cu+2<\/sup>(aq)<\/p>\n\n\n\n

Eo<\/sup>cell<\/sub>= 0.25 \u2013 0.34<\/p>\n\n\n\n

= -0.11v<\/p>\n\n\n\n

The reaction is not feasible (negative Eo<\/sup>cell<\/sub>)<\/p>\n\n\n\n

Cu(s) + 2 Ag+<\/sup>(aq) \u2192 Cu+2<\/sup>(aq) + 2 Ag(s)<\/p>\n\n\n\n

Eo<\/sup>cell <\/sub>=0.80 \u2013 0.34 = 0.46 v<\/p>\n\n\n\n

The reaction is feasible (positive Eo<\/sup>cell<\/sub>)<\/p>\n\n\n\n

Cu(s) + 2H+<\/sup>(aq)\u2192 Cu+2<\/sup> + H2 <\/sub>(g)<\/p>\n\n\n\n

Eo<\/sup>cell <\/sub>= 0 \u2013 0.34 = \u2013 0.34 v<\/p>\n\n\n\n

The reaction is not feasible (negative Eo<\/sup>cell<\/sub>)<\/p>\n\n\n\n

Fe(s) + 2H+<\/sup>(aq) \u2192 Fe+2<\/sup> (aq) + 3 H2<\/sub><\/p>\n\n\n\n

Eo<\/sup>cell <\/sub>= 0 \u2013 (-0.44) = 0.44 v<\/p>\n\n\n\n

The reaction is feasible (positive Eo<\/sup>cell<\/sub>)<\/p>\n\n\n\n

2.<\/strong> The emf of a cell corresponding to the reaction, Zn + 2H+ <\/sup>(aq) <\/strong>\u2192 Zn+2<\/sup> (0.1M) + H2<\/sub>(g) 1 atm is 0.30 v at 25 0<\/sup> C. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. (E0<\/sup>cell <\/sub>= \u2013 0.76 v )<\/strong><\/p>\n\n\n\n

Solution :<\/strong><\/p>\n\n\n\n

Eo<\/sup>cell<\/sub> = 0 \u2013 (-0.76) = 0.76 v<\/p>\n\n\n\n

Applying the Nernst equation,<\/p>\n\n\n\n

Ecell<\/sub> = Eo<\/sup>cell<\/sub> \u2013 0.0591 \/ 2 log [zN+2<\/sup>][H2] \/ [H]+<\/sup><\/p>\n\n\n\n

0.30 = 0.76 \u2013 0.0591 \/ 2 log 0.1 x 1 \/ [H+<\/sup>]2<\/sup><\/p>\n\n\n\n

log 0.1 \/ [H+<\/sup>]2 <\/sup>= 2 x 0.46 \/ 0.0591<\/p>\n\n\n\n

log 0.1 \u2013 log [H+<\/sup>]2<\/sup> = 15.56<\/p>\n\n\n\n

2 pH = 15.56 \u2013 log 0.1<\/p>\n\n\n\n

pH = 16.56 \/ 2 = 8.28<\/p>\n\n\n\n

3.<\/strong> The solution of metal of atomic mass X was electrolysed for 1 hour with a current of 0.25 ampere. The mass of the metal deposited was 0.295 g. Find the metal X if its valency is 2.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, I = 0.25 ampere, t= 1 hr = 60 x 60 = 3600s<\/p>\n\n\n\n

Q = I \u2715 t<\/p>\n\n\n\n

Q = 0.25 \u2715 3600<\/p>\n\n\n\n

= 900 coulombs<\/p>\n\n\n\n

Therefore, 900 coulombs of electricity deposit = 0.295<\/p>\n\n\n\n

96500 coulomb of electricity deposit = 0.295 X 96500 \/ 900 = 31.63g<\/p>\n\n\n\n

Valency of metal = atomic mass \/ equivalent mass<\/p>\n\n\n\n

Atomic mass of metal X = 31.63 X 2 = 63.26g<\/p>\n\n\n\n

Therefore, the metal X is copper.<\/p>\n","protected":false},"excerpt":{"rendered":"

The relationship between the quantity of electric charge passed through an electrolyte, and the amount of substance deposited at the electrodes was given by Faraday in 1834, in the form of the law of electrolysis. Faraday\u2019s First Law When an electric current is passed through an electrolyte, the amount of substance deposited is proportional to […]<\/p>\n","protected":false},"author":1,"featured_media":7072,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[736],"tags":[],"class_list":["post-7073","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-electrochemistry"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/12\/science-1.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/7073","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=7073"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/7073\/revisions"}],"predecessor-version":[{"id":7082,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/7073\/revisions\/7082"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/7072"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=7073"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=7073"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=7073"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}