JEE MAIN - Chemistry (2021 - 26th February Morning Shift - No. 20)
Consider the following reaction
$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$$.
The quantity of electricity required in Faraday to reduce five moles of $$MnO_4^ - $$ is ___________. (Integer answer)
$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$$.
The quantity of electricity required in Faraday to reduce five moles of $$MnO_4^ - $$ is ___________. (Integer answer)
Answer
25
Explanation
$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$$
1 mole of MnO4- required 5 moles of electrons or 5 F electricity.
$$ \therefore $$ 5 moles of MnO4- required 25 F electricity.
1 mole of MnO4- required 5 moles of electrons or 5 F electricity.
$$ \therefore $$ 5 moles of MnO4- required 25 F electricity.
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