The standard electrode potential of two half cells are given below-
$N i^{+ 2} + 2 e^{-} ⟶ N i; E^{0} = - 0.25 V$
$Z n^{+ 2} + 2 e^{-} ⟶ Z n; E^{0} = - 0.77 V$
The voltage cell formed by combining the two half cell would be:

-1.02 V

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+0.52 V

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+1.02 V

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-0.52 V

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Solution

The correct option is B +0.52 V
Given,
${N i}^{2 +} + 2 e^{-} ⟶ N i; E_{1}^{0} = - 0.25 V$
${Z n}^{2 +} + 2 e^{-} ⟶ Z n; E_{2}^{0} = - 0.77 V$
$E_{1}^{0} - E_{2}^{0} \Rightarrow$ ${N i}^{2 +} + 2 e^{-} + Z n \to N i + {Z n}^{2 +} + 2 e^{-}$
${N i}^{2 +} + Z n \to N i + {Z n}^{2 +}$
$E_{r \times n} = E_{1}^{0} - E_{2}^{0}$
$= (- 0.25) - (0.77)$
$= 0.52$ Volts
the voltage cell formed by combing two half cell would be $0.52$ Volts.

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Similar questions

Following are the values of standard potentials of two half cells, which of the following will be the value of standard cell potential E^o_cell:

Z n^{2 +} + 2 e^{-} \to Z n; E_{1} = - 7.62 V

F e^{2 +} + 2 e^{-} \to F e; E_{2} = - 7.81 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V

F e^{2 +} = Z n^{-} \to Z n^{2 +} + F e

F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V

F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

P b^{2 +} (a q) + 2 e^{-} \to P b (s); E^{0} = - 0.13 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

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