The electrode potential for Cu aq 2- e - Cu aq -and Cu aq - e - Cu s are -0-15 V and -0-50 V respectively- The value of E Cu 2- - Cu 0 will be-A- 0-150 VB- 0-500 VC- 0-325 VD- 0-65

The correct option is C 0.325 VWe know that: nE^0_half cell=n_1E^0_1+n_2E^0_2 Given data: Cu_(aq)^2++e^ → Cu_(aq)^+ n_1=1 & E^0_1=0.15V Cu_(aq)^++e^ → Cu ...

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Standard XII

Chemistry

Electrode Potential

The electrode...

Question

The electrode potential for
$C u_{(a q)}^{2 +} + e^{-} \to C u_{(a q)}^{+}$ and $C u_{(a q)}^{+} + e^{-} \to C u_{(s)}$ are +0.15 V and +0.50 V respectively. The value of $E_{C u^{2 +} / C u}^{o}$ will be:

0.150 V

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0.500 V

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0.325 V

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0.65

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Solution

The correct option is C 0.325 V
We know that:
$n E_{h a l f c e l l}^{0} = n_{1} E_{1}^{0} + n_{2} E_{2}^{0}$
Given data:
$C u_{(a q)}^{2 +} + e^{-} \to C u_{(a q)}^{+}$
$n_{1} = 1 & E_{1}^{0} = 0.15 V$
$C u_{(a q)}^{+} + e^{-} \to C u_{(s)}$
$n_{2} = 1 & E_{2}^{0} = 0.50$
We have to calculate $E^{0}$ for
$C u^{2 +} + 2 e^{-} \to C u (s)$
$n = 2$
On putting all values we get
$2 E_{C u^{2 +} / C u}^{0} = 0.15 + 0.50$
$E_{C u^{2 +} / C u}^{0} = 0.325 V$

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Similar questions

Q. The electrode potentials for

C u_{(a q)}^{2 +} + e^{-} \to C u_{(a q)}^{+}

and

C u_{(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}^{\circ}

will be

Q. The electrode potentials for :

C u^{2 +} (a q) + e^{-} \to C u^{+} (a q)

and

C u^{+} (a q) + e^{-} \to C u (s)

are +0.15V and +0.50V respectively. The value of

E_{C u^{2 +} / C u}^{\circ}

will be:

Q. Give Nernst equation:
Calculate the electrode potential of the following single electrode.

C u_{(a q)}^{+ +} (C = 0.01 M) / C u; (E^{\circ} = + 0.337 V)

The $E^{\circ}$ for $C u^{2 +} / C u^{+}$ ; $C u^{+} / C u$ , $C u^{2 +} / C u$ are 0.15V, 0.50 V and 0.325 V respectively. The redox cell showing redox reaction $2 C u^{+} \to C u^{2 +} + C u$ is made. The $E^{\circ}$ of this cell reaction and $Δ G^{\circ}$ may be:

Q. Calculate the equailibrium constant for the reaction:

{C u}_{(s)} + {2 A g}_{(a q)}^{+} ⟶ {C u}_{(a q)}^{2 +} + {2 A g}_{(s)}

Given that :

E_{A g^{+} / A g}^{0} = 0.80 V, E_{C u^{2 +} / C u}^{0} = 0.34 V .

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The electrode potential for Cu2+(aq)+e−→Cu+(aq) and Cu+(aq)+e−→Cu(s) are +0.15 V and +0.50 V respectively. The value of EoCu2+/Cu will be:

The correct option is C 0.325 VWe know that: nE0half cell=n1E01+n2E02 Given data: Cu2+(aq)+e−→Cu+(aq) n1=1 & E01=0.15V Cu+(aq)+e−→Cu(s) n2=1 & E02=0.50 We have to calculate E0 for Cu2++2e−→Cu(s) n=2 On putting all values we get 2E0Cu2+/Cu=0.15+0.50 E0Cu2+/Cu=0.325V

The electrode potential for
$C u_{(a q)}^{2 +} + e^{-} \to C u_{(a q)}^{+}$ and $C u_{(a q)}^{+} + e^{-} \to C u_{(s)}$ are +0.15 V and +0.50 V respectively. The value of $E_{C u^{2 +} / C u}^{o}$ will be: