SATHEE: Redox Reactions And Electrochemistry Ques 630

Redox Reactions And Electrochemistry Ques 630

Question: Given $E {^{\circ}}_{C u^{2 +} / C u} = 0.34 V,, E {^{\circ}}_{C u^{2 +} / C u} = 0.15 V$ Standard electrode potential for the half cell $C u^{+} / C u$ is

Options:

A) 0.38 V

B) 0.53 V

C) 0.19 V

D) 0.49 V

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Answer:

Correct Answer: B

Solution:

[b] $C u^{2 +} + e^{-} \overset{}{\to} C u^{+};$

$E_{1}^{^{\circ}} = 0.15 V; Δ G_{1}^{^{\circ}} = - n_{1} E_{1}^{^{\circ}} F$

$C u^{2 +} 2 e \overset{}{\to} C u;$

$E_{2}^{^{\circ}} = 0.34 V; Δ G_{2}^{^{\circ}} = - n_{2} E_{2}^{^{\circ}} F$

On subracting eq. (i) from eq. (ii) we get

$C u^{+} + e^{-} \overset{}{\to} C u; Δ G_{2}^{^{\circ}} = Δ G^{\circ} = Δ G_{2}^{^{\circ}} - Δ G_{1}^{^{\circ}}$ $- n E^{\circ} F = - (n_{2} E^{\circ} F - n_{1} E_{1}^{^{\circ}} F)$ $E^{\circ} = \frac{n_{2} E_{2}^{^{\circ}} F - n_{1} E_{1}^{^{\circ}} F}{n F}$ $= \frac{2 \times 0.34 - 0.15}{1} = 0.53 V$

Redox Reactions And Electrochemistry Ques 630

Question: Given $E {^{\circ}}_{C u^{2 +} / C u} = 0.34 V,, E {^{\circ}}_{C u^{2 +} / C u} = 0.15 V$ Standard electrode potential for the half cell $C u^{+} / C u$ is

Options:

Answer:

Solution:

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Redox Reactions And Electrochemistry Ques 630

Question: Given E∘Cu2+/Cu=0.34V,,E∘Cu2+/Cu=0.15V Standard electrode potential for the half cell Cu+/Cu is

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Question: Given $E {^{\circ}}_{C u^{2 +} / C u} = 0.34 V,, E {^{\circ}}_{C u^{2 +} / C u} = 0.15 V$ Standard electrode potential for the half cell $C u^{+} / C u$ is