JEE Advance - Chemistry (1993 - No. 18)
The standard reduction potential for the half-cell
$$NO_3^-$$ + 2H+ (aq) + e $$\to$$ NO2 (g) + H2O is 0.78 V
(i) Calculate the reduction potential in 8 M H+
(ii) What will be the reduction potential of the half-cell in a neutral solution? Assume all the other species to be at unit concentration.
$$NO_3^-$$ + 2H+ (aq) + e $$\to$$ NO2 (g) + H2O is 0.78 V
(i) Calculate the reduction potential in 8 M H+
(ii) What will be the reduction potential of the half-cell in a neutral solution? Assume all the other species to be at unit concentration.
(i) 0.887 V, (ii) -0.046 V
(i) 0.78 V, (ii) 0.0 V
(i) 0.90 V, (ii) -0.10 V
(i) 0.80 V, (ii) -0.05 V
(i) 0.85 V, (ii) -0.04 V
Explanation
Reaction :
$$NO_3^ - (aq) + 2{H^ + }(aq) + e \to N{O_2}(g) + {H_2}O(l)$$
(1) $${E_{cell}} = E_{cell}^0 - {{0.059} \over n}\log {{[N{O_2}][{H_2}O]} \over {[NO_3^ - ]{{[{H^ + }]}^2}}}$$
$$ = 0.78 - {{0.059} \over 1}\log {{1 \times 1} \over {1 \times {{(8)}^2}}} = 0.887$$ V
(2) In neutral solution, [H+] = 10$$-$$7 M
$$\therefore$$ $$E = 0.78 - {{0.059} \over 1}\log {{1 \times 1} \over {1 \times {{({{10}^{ - 7}})}^2}}} = - 0.046$$ V
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