Consider the following Galvanic cell. By what | vedclass

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(C) The cell reaction is $H_{2}(g) + Cl_{2}(g) \rightarrow 2H^{+}(aq) + 2Cl^{-}(aq)$.The Nernst equation for the cell potential at $298 \ K$ is $E_{ce

Vedclass · Accepted Answer

$-0.1182 \ V$

Vedclass · Answer

(C) The cell reaction is $H_{2}(g) + Cl_{2}(g) ightarrow 2H^{+}(aq) + 2Cl^{-}(aq)$.The Nernst equation for the cell potential at $298 \ K$ is $E_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \log \frac{[H^{+}]^{2} [Cl^{-}]^{2}}{P_{H_{2}} P_{Cl_{2}}}$.When the concentration of ions in both compartments is increased by a factor of $10$,the new concentrations are $[H^{+}]' = 10[H^{+}]$ and $[Cl^{-}]' = 10[Cl^{-}]$.The new cell potential $E'_{cell}$ is given by:$E'_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \log \frac{(10[H^{+}])^{2} (10[Cl^{-}])^{2}}{P_{H_{2}} P_{Cl_{2}}}$$E'_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \log \left( 10^{4} 	imes \frac{[H^{+}]^{2} [Cl^{-}]^{2}}{P_{H_{2}} P_{Cl_{2}}} ight)$$E'_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \left( \log \frac{[H^{+}]^{2} [Cl^{-}]^{2}}{P_{H_{2}} P_{Cl_{2}}} + \log 10^{4} ight)$$E'_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \log \frac{[H^{+}]^{2} [Cl^{-}]^{2}}{P_{H_{2}} P_{Cl_{2}}} - \frac{0.0591}{2} 	imes 4$$E'_{cell} = E_{cell} - 0.1182 \ V$.The change in cell potential is $\Delta E = E'_{cell} - E_{cell} = -0.1182 \ V$.

Consider the following Galvanic cell. By what value does the cell voltage change when the concentration of ions in both the anodic and cathodic compartments is increased by a factor of $10$ at $298 \ K?$

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