Pt | H_2 (1 atm) | H^{+} (0.001 M) || H^{+} | vedclass

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Question

(A) The given cell is a concentration cell.For a hydrogen concentration cell,the $E_{cell}$ is given by the Nernst equation:$E_{cell} = E_{cell}^{\cir

Vedclass · Accepted Answer

$0.1182$

Vedclass · Answer

(A) The given cell is a concentration cell.For a hydrogen concentration cell,the $E_{cell}$ is given by the Nernst equation:$E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log \frac{[H^{+}]_{anode}}{[H^{+}]_{cathode}}$Since $E_{cell}^{\circ} = 0$ for a concentration cell and $n = 1$ for the reaction $H^{+} + e^{-} ightarrow \frac{1}{2} H_2$:$E_{cell} = -0.0591 \log \frac{0.001}{0.1}$$E_{cell} = -0.0591 \log (10^{-2})$$E_{cell} = -0.0591 	imes (-2) = 0.1182 \ V$

$Pt | H_2 (1 \ atm) | H^{+} (0.001 \ M) || H^{+} (0.1 \ M) | H_2 (1 \ atm) | Pt$. What will be the value of $E_{cell}$ for this cell? ............. $V$

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