: Can 1M NiSO4 be stored in a vessel made up of copper ? Given E°Ni2+/Ni = -1.02V and E°Cu2+/Cu = 0.34 V how to decide which is anode shouldn't the one with lesser reduction potential be anode which in this case should be negative value

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Can 1M $N i S O_{4}$ be stored in a vessel of copper? Given that $E_{\frac{{N i}_{2 +}}{N i}}^{0}$ = -1.20 V and E°Cu2+/Cu =0.34V

Q. Calculate the

E

and

E^{\circ}

of the cell

N i | N i^{2 +} | | C u^{2 +} | C u

from the following half-cell reactions:

N i^{2 +} + 2 e^{-} \to N i; E^{\circ} = - 0.25 v o l t

C u^{2 +} + 2 e^{-} \to C u; E^{\circ} = + 0.34 v o l t

(Given:

[N i^{2 +}] = 1 M

and

[C u^{2 +}] = 10^{- 3} M)

Q. During electrolysis,

O_{2} (g)

is evolved at anode in

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) C u (s) ⟶ & C u^{2 +} (a q) + 2 e^{-} E_{C u | C u^{2 +}}^{⊝} = - 0.34 V (w r o n g) 2 H_{2} O (l) ⟶ & O_{2} (g) + 4 H^{\oplus} (a q) + 4 e^{-} E_{H_{2} O | O_{2}}^{⊝} = - 1.23 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

\begin{matrix} I n (d) \end{matrix}

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) N i (s) ⟶ & N i^{2 +} (a q) + 2 e^{-} E_{N i | N i^{2 +}}^{⊝} = - 0.25 V (w r o n g) 4 ⊝ O H (a q) ⟶ & O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} E_{⊝ O H | O_{2}}^{\oplus} = - 0.4 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

N i | N i^{2 +} | | C u^{2 +} | C u

The standard EMF of the above cell is 0.59 V. The standard electrode potential (reduction potential) of the copper electrode is 0.34 V. Calculate the standard electrode potential of nickel electrode.

Given:
$E_{A g^{+} / A g}^{0} = 0.80 V$ ,
$E_{M g^{2 +} / M g}^{0} = - 2.37 V$ ,
$E_{C u^{2 +} / C u}^{0} = 0.34 V$ ,
$E_{H g^{2 +} / H g}^{0} = 0.79 V$
Which of the following statements is/are correct ?

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