For the cell
Cu(s) | Cu2+ (aq) (0.1 M) || Ag+(aq) (0.01 M) | Ag(s)
the cell potential E1 = 0.3095 V
For the cell
Cu(s) | Cu2+ (aq) (0.01 M) || Ag+(aq) (0.001 M) | Ag(s)
the cell potential = ____________ $\times$ 10$-$2 V. (Round off the nearest integer).
[Use : ${{2.303RT} \over F}$ = 0.059]
Answer (integer)
28
Solution
Cell reaction is :<br><br>$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$<br><br>Now, $${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$$ .... (1)<br><br>$\therefore$ $${E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}}$$ ....(2)<br><br>From (1) and (2), E<sub>2</sub> = 0.28 V = 28 $\times$ 10<sup>$-$2</sup> V
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
This question is part of PrepWiser's free JEE Main question bank. 127 more solved questions on Electrochemistry are available — start with the harder ones if your accuracy is >70%.