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Standard XII

Chemistry

Electrode Potential

A voltaic cel...

Question

A voltaic cell in set up at $25^{o} C$ with the following half cells:
$A l / A l^{3 +} (0.001 M)$ and $N i / N i^{2 +} (0.50 M)$
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
$E_{N i^{2 +} / N i} = - 0.25 V a n d E_{A l^{3 +} / A l}^{o} = - 1.66 V (l o g (8 \times 10^{- 6}) = - 0.54)$

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Solution

$A l^{+ 3} + 3 e^{-} \to A l$
$E_{0} = - 1.66 V$
$N i^{2 +} + 2 e^{-} \to N i$
$E_{0} = - 0.25 V$
$E_{c e l l}^{o} = - 0.25 - (- 1.66) = 1.41 V$
Thus, aluminium electrode is anode and nickel electrode is cathode reaction.
$3 N i^{+ 2} + 2 A l \to 3 N i + 2 A l^{+ 3}$
$E^{o} = E_{c e l l}^{o} - \frac{0.0591}{6} l o g \frac{(1 \times 10^{- 3})^{2}}{(5 \times 10^{- 1})^{3}}$
$= 1.41 + 0.005319 = 1.415 V$

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Similar questions

Q. Determine the standard emf of the cell and standard free energy change of the cell reaction

Z n | Z n^{2 +} | | N i^{2 +} | N i

. The standard reduction potentials of

Z n^{2 +} | Z n

and

N i^{2 +} | N i

half cells are

- 0.76 V

and

- 0.25 V

respectively.

Q. The standard reduction potential of the half cells at

298 K

are,

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{o} = - 0.28 V

M g^{2 +} (a q) + 2 e^{-} \to M g (s); E^{o} = - 2.37 V

The standard cell potential for a voltaic cell constructed using the two half reaction is

Q. Derive Nernst equation for the cell.

N i (s) | N i^{2 +} (a q . 0.1 M) | | A g^{+} (a q . 0.1 M | A g (s)

and also find its cell potential. Given:

E_{A g^{+} / A g}^{\circ} = 0.80 v o l t

and

E_{N i^{2 +} / N i}^{\circ} = - 0.25 v o l t

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)

Construct a cell from ${N i}^{2 +} ∣ ∣ N i a n d {C u}^{2 +} ∣ ∣ C u$ half cells. Write the cell reaction and calculate
the standard of the cell. ${E^{0}}_{N i} = - 0.236 V$ and ${E^{0}}_{C u} = 0.377$

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A voltaic cell in set up at 25oC with the following half cells:Al/Al3+(0.001M) and Ni/Ni2+(0.50M)Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential. ENi2+/Ni=−0.25V and EoAl3+/Al=−1.66V (log(8×10−6)=−0.54)

Al+3+3e−→Al E0=−1.66V Ni2++2e−→NiE0=−0.25VEocell=−0.25−(−1.66)=1.41 VThus, aluminium electrode is anode and nickel electrode is cathode reaction. 3Ni+2+2Al→3Ni+2Al+3Eo=Eocell−0.05916log(1×10−3)2(5×10−1)3=1.41+0.005319=1.415 V