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Electrolysis and Electrolytes

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Question

A solution containing equimolar amounts of $N i C l_{2}$ and $S n B r_{2}$ is electrolysed using a 9V battery and graphite electrodes. What are the first products formed?
Half reaction $N i^{2 +} (a q) + 2 e^{-} \to N i (s)$ Standard reduction potential (V) -0.236
Half reaction $S n^{2 +} (a q) + 2 e^{-} \to S n (s)$ Standard reduction potential (V) -0.141
Half reaction $B r_{2} (a q) + 2 e^{-} \to 2 B r^{-} (a q)$ Standard reduction potential (V) 1.077
Half reaction $C l_{2} (a q) + 2 e^{-} \to 2 C l^{-} (a q)$ Standard reduction potential (V) 1.360

Ni (s) at cathode,

C l_{2}

(aq) at anode

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Ni (s) at cathode,

B r_{2}

(aq) at anode

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Sn (s) at cathode,

B r_{2}

(aq) at anode

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Sn (s) at cathode,

C l_{2}

(aq) at anode

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Solution

The correct option is C
Sn (s) at cathode, $B r_{2}$ (aq) at anode

The ion having higher reduction potential will deposit first at the cathode
Among $N i^{+ 2}$ and $S n^{+ 2}$ , $S n^{+ 2}$ has higher reduction potential so, it will be deposit at the cathode.
Half cell reaction at cathode:
$S n^{+ 2} + 2 e^{-} \to S n (s)$

$B r^{-}$ has lower reduction potential or higher oxidation potential than $C l^{-}$
So the half cell reaction at anode is:
$2 B r^{-} \to B r_{2} + 2 e^{-}$
Hence, the correct option is C.

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Similar questions

Q. Assertion: Starting from aq. copper [II] sulphate solution, equations for the reactions at the cathode and anode during electrolysis of aq.

C u S O_{4}

using active copper electrodes is;
at anode:

C u^{+ 2} + 2 e^{-} \to C u

at cathode:

C u \to C u^{+ 2} + 2 e^{-}

Reason: Oxidation always take place at anode and reduction at cathode.

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

At 298 K, the standard reduction potentials for the following half-reactions are given. Which acts as an anode with others in an electrochemical cell?

$Z n^{+ 2} (a q) + 2 e^{-} \to Z n (s),$ -0.762 V

$C r^{+ 3} (a q) + 3 e^{-} \to C r (s),$ -0.740 V

$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g),$ 0.000 V

$F e^{+ 3} (a q) + e^{-} \to F e^{2 +} (a q),$ +0.762 V

Q. Standard reduction potentials of the half reactions are given below

F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q); E^{0} = + 2.85 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{0} = + 1.36 V

B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q); E^{0} = + 1.06 V

I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q); E^{0} = + 0.53 V

The strongest oxidising and reducing agents respectively are:

Q. The standard reduction potential at

298

K for the following half cell reaction:

Z n_{(a q)}^{2 +} + 2 e \to Z n_{(s)} E^{o} = - 0.762

C r_{(a q)}^{3 +} + 3 e \to C r_{(s)} E^{o} = 0.740

2 H_{(a q)}^{+} + 2 e \to H_{2 (g)} E^{o} = 0.0

F_{2 (q)} + 2 e \to 2 F_{(a q)}^{-} E^{o} = 2.87

Which of the following is the strongest reducing agent?

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