MnO 2 is prepared by electrolysis of aqueous solution of MnSO 4, as per reaction Mn 2+(a q)+2 H 2 O arrow MnO 2(s)+2 H (a q)++ H 2(g) Passing a current of 25 A for 30 hours gives one kg of MnO 2 . What is the current efficiency? (Mol. wt. of MnO 2 =87)

Question

MnO 2 is prepared by electrolysis of aqueous solution of MnSO 4, as per reaction Mn 2+(a q)+2 H 2 O arrow MnO 2(s)+2 H (a q)++ H 2(g) Passing a current of 25 A for 30 hours gives one kg of MnO 2 . What is the current efficiency? (Mol. wt. of MnO 2 =87) (A) 20.54 % (B) 25 % (C) 49.2 % (D) 82.16 %

Tardigrade Admin · Accepted Answer

Applying, w=Z × I × t w=1 kg =1000 g , I= current required, t=30 hr =30 × 3600 s I=(1000 × 2 × 96500/87) × (1/30 × 3600)=20.54 ampere Current efficiency =(20.54/25) × 100=82.16 %

$20.54%$

$25%$

$49.2%$

$82.16%$

Applying, $w = Z \times I \times t$

$w = 1 k g = 1000 g, I =$ current required, $t = 30 h r = 30 \times 3600 s$

$I = \frac{1000 \times 2 \times 96500}{87} \times \frac{1}{30 \times 3600} = 20.54$ ampere

Current efficiency $= \frac{20.54}{25} \times 100 = 82.16%$

Q. MnO2​ is prepared by electrolysis of aqueous solution of MnSO4​, as per reaction Mn(aq)2+​+2H2​O→MnO2(s)​+2H(aq)+​+H2(g)​ Passing a current of 25A for 30 hours gives one kg of MnO2​. What is the current efficiency? (Mol. wt. of MnO2​ =87)

20.54%

25%

49.2%

82.16%

Applying, w=Z×I×t w=1kg=1000g,I= current required, t=30hr=30×3600s I=871000×2×96500​×30×36001​=20.54 ampere Current efficiency =2520.54​×100=82.16%

$20.54%$

$25%$

$49.2%$

$82.16%$

Applying, $w = Z \times I \times t$

$w = 1 k g = 1000 g, I =$ current required, $t = 30 h r = 30 \times 3600 s$

$I = \frac{1000 \times 2 \times 96500}{87} \times \frac{1}{30 \times 3600} = 20.54$ ampere

Current efficiency $= \frac{20.54}{25} \times 100 = 82.16%$