JEE MAIN - Chemistry (2021 - 27th July Evening Shift - No. 22)
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]
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
28
Explanation
Cell reaction is :
$$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$$
Now, $${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$$ .... (1)
$$\therefore$$ $${E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}}$$ ....(2)
From (1) and (2), E2 = 0.28 V = 28 $$\times$$ 10$$-$$2 V
$$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$$
Now, $${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$$ .... (1)
$$\therefore$$ $${E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}}$$ ....(2)
From (1) and (2), E2 = 0.28 V = 28 $$\times$$ 10$$-$$2 V
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