Verified Solution • 2025–26 Session moderate

QJEEM21Chd6db: JEE Main 2021 Chemistry - Class 12 JEEM21Chd6db Solution: For the cell Cu(s)

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 : 2.303RTF{{2.303RT} \over F} = 0.059]

Step-by-Step Verified Solution

Cell reaction is :

Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)

Now, Ecell=Ecello−0.0592log⁡[Cu2+][Ag+]2{E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}} .... (1)

∴\therefore E1=0.3095=Ecello−0.0592.log⁡0.01(0.001)2{E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}} ....(2)

From (1) and (2), E2 = 0.28 V = 28 ×\times 10 −- 2 V

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