\(Pt(s) | H_{2_{(g)}} (1 bar) | H^{+}_{(aq)}(1M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt(s)\)
The Ecell for the given cell is 0.1115 V at 298 K when \(\frac{[M^+(aq)}{[M^{3+}(aq)]}=10^{a}\)
The value of a is _________
Given: \(E^{\theta}_{M^{3+}/M^+}=0.2\;V\)
\(\frac{2.303RT}{F}=0.059V\)
The Ecell for the given cell is 0.1115 V at 298 K when \(\frac{[M^+(aq)}{[M^{3+}(aq)]}=10^{a}\)
The value of a is _________
Given: \(E^{\theta}_{M^{3+}/M^+}=0.2\;V\)
\(\frac{2.303RT}{F}=0.059V\)
Correct Answer: 3
Solution and Explanation
Top Questions on Electrochemistry
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)- CBSE CLASS XII - 2025
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- (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: \[ \text{Sn}(s) \mid \text{Sn}^{2+} (\text{0.004 M}) \parallel \text{H}^+ (\text{0.02 M}) \mid \text{H}_2 (\text{1 Bar}) \mid \text{Pt}(s) \] (Given: \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V}, E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \, \text{V} \)) \vspace{5pt}
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\( \Lambda_m^0 (\text{HC}) = 426 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
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\( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).- CBSE CLASS XII - 2025
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