JEE Advance - Chemistry (1996 - No. 13)
The standard reduction potential for Cu2+|Cu is +0.34 V. Calculate the reduction potential at pH = 14 for the above couple. Ksp of Cu(OH)2 is 1.0 $$\times$$ 10-19
-0.22 V
+0.22 V
+0.34 V
-0.34 V
0.00 V
Explanation
Given, $$E_{C{u^{2 + }}|Cu}^0(aq) = + 0.34V$$
$$Cu{(OH)_2}(s) \to C{u^{2 + }}(aq) + 2O{H^ - }(aq)$$, $${K_{sp}} = [C{u^{2 + }}]{[O{H^ - }]^2}$$
Given, pH = 14, [H+] = 10$$-$$14 M, [OH$$-$$] = 1 M
Ksp of $$Cu{(OH)_2} = 1.0 \times {10^{ - 19}}$$
$$\therefore$$ $$1.0 \times {10^{ - 19}} = [C{u^{2 + }}]{[O{H^ - }]^2}$$
or, $$[C{u^{2 + }}]{[1]^2} = 1.0 \times {10^{ - 19}}$$ [$$\because$$ $$C{u^{2 + }} + 2e \to Cu$$]
$$\therefore$$ $$[C{u^{2 + }}] = {10^{ - 19}}M$$
$$\therefore$$ $${E_{C{u^{2 + }}|Cu}} = E_{C{u^{2 + }}|Cu}^0 - {{0.059} \over 2}\log {1 \over {[C{u^{2 + }}]}}$$
$$\therefore$$ $${E_{C{u^{2 + }}|Cu}} = 0.34 - {{0.059} \over 2}\log {1 \over {({{10}^{ - 19}})}} = - 0.22V$$
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