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

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$-0.1182 \ V$

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(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 $298 \ K?$

Similar Questions

$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

What minimum decomposition potential is necessary to produce $Cl_2$ gas in the following reaction?
Given: $(\frac{2.303RT}{F} = 0.06)$
$Sn^{+2} (1 \ M) + 2Cl^{-} (2 \ M) \rightleftharpoons Sn_{(s)} + Cl_2 (1 \ atm)$
Given: $E^{o}_{Sn^{+2}/Sn} = -0.14 \ V$,$E^{o}_{Cl_2/Cl^{-}} = 1.4 \ V$

Calculate the $pH$ of an $HCl$ solution at $298 \ K$ for the following cell:
$Pt_{(s)} \mid H_2 \ (1 \ bar) \mid HCl \ (xM) \parallel Ag^+ \ (0.01 \ M) \mid Ag_{(s)}$
Given that the standard cell potential $E^\circ_{cell} = 1.05 \ V$. (in $.73$)

In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?
$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$

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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 $298 \ K?$

Similar Questions

$EMF$ of the following concentration cell will be ............. $V$$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

What minimum decomposition potential is necessary to produce $Cl_2$ gas in the following reaction? Given: $(\frac{2.303RT}{F} = 0.06)$ $Sn^{+2} (1 \ M) + 2Cl^{-} (2 \ M) \rightleftharpoons Sn_{(s)} + Cl_2 (1 \ atm)$ Given: $E^{o}_{Sn^{+2}/Sn} = -0.14 \ V$,$E^{o}_{Cl_2/Cl^{-}} = 1.4 \ V$

Calculate the $pH$ of an $HCl$ solution at $298 \ K$ for the following cell:$Pt_{(s)} \mid H_2 \ (1 \ bar) \mid HCl \ (xM) \parallel Ag^+ \ (0.01 \ M) \mid Ag_{(s)}$Given that the standard cell potential $E^\circ_{cell} = 1.05 \ V$. (in $.73$)

In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$