(A) The given cell is composed of a copper electrode as the anode and a silver electrode as the cathode.The standard cell potential is given by the fo

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(A) The given cell is composed of a copper electrode as the anode and a silver electrode as the cathode.The standard cell potential is given by the formula:$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$Substituting the given values:$E^{\circ}_{cell} = E^{\circ}_{Ag^{+}|Ag} - E^{\circ}_{Cu^{2+}|Cu}$$0.647 \ V = E^{\circ}_{Ag^{+}|Ag} - 0.153 \ V$$E^{\circ}_{Ag^{+}|Ag} = 0.647 \ V + 0.153 \ V = 0.8 \ V$

Class 12 Chemistry Electrochemistry Electrode potential and ECell

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The $E^{\circ}_{cell}$ of $Cu_{(s)} | Cu^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ is $0.647 \ V$. Calculate the $E^{\circ}_{Ag}$ if $E^{\circ}_{Cu}$ is $0.153 \ V$. (in $V$)

MHT CET 2024 All MHT CET

A
$0.8$
B
$0.5$
C
$-0.8$
D
$-0.5$

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Electrochemistry

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Electrode potential and ECell

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The $E^{\circ}_{cell}$ of $Cu_{(s)} | Cu^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ is $0.647 \ V$. Calculate the $E^{\circ}_{Ag}$ if $E^{\circ}_{Cu}$ is $0.153 \ V$. (in $V$)

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Calculate $E_{\text{cell}}^{\circ}$ for $Cd_{(s)}|Cd^{2+}_{(1M)}||Ag^{+}_{(1M)}|Ag_{(s)}$. Given: $E^{\circ}_{Cd^{2+}/Cd} = -0.403 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.799 \ V$. (in $V$)

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