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Electrochemical Series

Calculate the...

Question

Calculate the reduction potential for the following half cell at $25^{o} C$ .

$C u (s) | C u^{2 +} (a q, 0.2 M); E_{C u^{2 +} / C u}^{0} = 0.34 V$

Take :
[ $l o g_{10} 5 = 0.7$ ]

A

E_{C u^{2 +} / C u} = 0.32 V

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B

E_{C u^{2 +} / C u} = 0.39 V

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C

E_{C u^{2 +} / C u} = 0.50 V

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D

E_{C u^{2 +} / C u} = - 0.2 V

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Solution

The correct option is A $E_{C u^{2 +} / C u} = 0.32 V$
$C u (s) | C u^{2 +} (a q, 0.2 M); E_{C u^{2 +} / C u}^{0} = 0.34 V$

$E_{C u^{2 +} / C u} = E_{C u^{2 +} / C u}^{0} - \frac{2.303 R T}{n F} l o g \frac{[M (s)]}{[M^{n +} (a q)]}$

$n = 2$

$E_{C u^{2 +} / C u} = 0.34 - \frac{0.059}{2} l o g \frac{1}{[C u^{2 +}]}$

$E_{C u^{2 +} / C u} = 0.34 - \frac{0.059}{2} l o g \frac{1}{[0.2]}$

$E_{C u^{2 +} / C u} = 0.34 - \frac{0.059}{2} \times l o g 5$

$E_{C u^{2 +} / C u} = 0.34 - \frac{0.059}{2} \times 0.7 \approx 0.32 V$

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Similar questions

Q. Calculate the cell potential of the following cell at

25

^{0}

P t, H_{2 (1 a t m)} | H {^{+}}_{(0.01 M)} | | C u {^{+ 2}}_{(0.1 M)} | C u

[E {^{0}}_{C u^{2 +} / C u} = 0.337 V]

Q. The standard reduction potentials for

C u^{2 +} / C u

C u^{2 +} / C u^{+}

0.337 V

0.153 V

respectively. The standard electrode potential of

C u^{+} / C u

Q. The electrode potentials for

C u^{2 +} (a q) + e^{-} \to C u^{+} (a q) and C u^{2 +} (a q) + e^{-} \to C u (s) are + 0.15 V and + 0.50 V

respectively. The value of

E_{C u^{2 +} / C u}^{0}

Q. The standard reduction potentials of

{C u}^{2 +} / C u

{C u}^{2 +} / {C u}^{+}

0.337

0.153

volts resectively. The standard electrode potential for

{C u}^{+} / C u

half cell will be:

E^{\circ}

values of following reactions:

C u_{(s)} + Z n_{(a q)}^{2 +} \to Z n_{(s)} + C u_{(a q)}^{2 +}

E_{C u^{2 +} / C u}^{0} = 0.34 V, E_{Z n^{2 +} / Z n}^{0} = - 0.76 V

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Electrochemical Series

Standard VII Chemistry

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