The potential for the given half cell at 298K is -............ ×10^(-2)V. 2H(aq)++2e^(-)→H_2( g) H^(+)=1M,P_H2=2 atm (Given2.303 RT/F=0.06 V, log2=0.3)

Correct answer: 1. The Nernst equation for the given cell reaction 2H^(+)+2e^(-)→H_2 is E=E^(°)-0.06/2×log p_H2H+^2 E=0-0.06/2log21^2 E=-0.03×0.3=-0.009=-9×10^(-3) E=-0.9×10^(-2)V

Q20 / 90 · JEE Main 2024 (01 Feb Shift 1) · Chemistry

The potential for the given half cell at $298 K$ is $(-)$ ............ $\times 10^{- 2} V .$
$2 H_{(aq)}^{+} + 2 e^{-} \to H_{2} (g)$
$[H^{+}] = 1 M, P_{H_{2}} = 2 atm$
(Given $2.303 RT / F = 0.06 V, \log 2 = 0.3$ )

Correct answer: 1

Solution

The Nernst equation for the given cell reaction $2 H^{+} + 2 e^{-} \to H_{2}$ is

$E = E^{°} - \frac{0.06}{2} \times \log \frac{p_{H_{2}}}{{[H^{+}]}^{2}}$

$E = 0 - \frac{0.06}{2} \log \frac{2}{{[1]}^{2}}$

$E = - 0.03 \times 0.3 = - 0.009 = - 9 \times 10^{- 3}$

$E = - 0.9 \times 10^{- 2} V$

Hence, the correct answer is 1.

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The potential for the given half cell at 298K is -............ ×10-2V.2H(aq)++2e-→H2( g)H+=1M,PH2=2 atm(Given2.303 RT/F=0.06 V, log2=0.3)

Solution

Relevant Questions

The potential for the given half cell at $298 K$ is $(-)$ ............ $\times 10^{- 2} V .$
$2 H_{(aq)}^{+} + 2 e^{-} \to H_{2} (g)$
$[H^{+}] = 1 M, P_{H_{2}} = 2 atm$
(Given $2.303 RT / F = 0.06 V, \log 2 = 0.3$ )