How many coulombs of electricity are required for reduction of 1 mole of MnO4- to Mn2+ ?

Question

How many coulombs of electricity are required for reduction of 1 mole of MnO4- to Mn2+ ? (A) 96500 C (B)  9.65× 106C (C)  4.83× 105C (D)  1.93× 105C

Tardigrade Admin · Accepted Answer

underset1 text mol MnO 4-1+8 H ++ underset5 mole 5 e- longrightarrow Mn 2++4 H 2 O ie, 5 F Hence, electricity required =5 × 96500 C =482500 C =4.83 × 105 C

Q. How many coulombs of electricity are required for reduction of $1$ mole of $M n O_{4}^{-}$ to $M n^{2 +}$ ?

96500 C

$9.65 \times 1 0^{6} C$

$4.83 \times 1 0^{5} C$

$1.93 \times 1 0^{5} C$

$1 mol M n O_{4}^{- 1} + 8 H^{+} + 5 m o l e 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
ie, $5 F$
Hence, electricity required
$= 5 \times 96500 C$
$= 482500 C$
$= 4.83 \times 1 0^{5} C$

Q. How many coulombs of electricity are required for reduction of 1 mole of MnO4−​ to Mn2+ ?

96500 C

9.65×106C

4.83×105C

1.93×105C

1 mol MnO4−1​​+8H++5mole5e−​⟶Mn2++4H2​O ie, 5F Hence, electricity required =5×96500C =482500C =4.83×105C

Q. How many coulombs of electricity are required for reduction of $1$ mole of $M n O_{4}^{-}$ to $M n^{2 +}$ ?

$9.65 \times 1 0^{6} C$

$4.83 \times 1 0^{5} C$

$1.93 \times 1 0^{5} C$

$1 mol M n O_{4}^{- 1} + 8 H^{+} + 5 m o l e 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
ie, $5 F$
Hence, electricity required
$= 5 \times 96500 C$
$= 482500 C$
$= 4.83 \times 1 0^{5} C$