Electrolysis of a solution of MnSO₄ in aqueous sulphuric acid is a method for preparation of MnO2. Passing a current of 27A for 24 hrs gives 1 kg of MnO2. The current efficiency of this process is

Dear student,
Please find the solution to the asked query:
The reactions taking place in the question:
At anode: Mn²⁺(aq) + 2H₂O(l)$\to$MnO₂ (s) + 4H⁺ (aq) + 2e
At cathode: 2H⁺ (aq) +2e $\to$H₂(g)

Overall reaction: Mn²⁺(aq) +2H₂O(l) $\to$ MnO₂(s) +2H⁺(aq) +H₂(g)
Amount of  current is given by 
$I =m\frac{F\times Z}{t\times M}$
where m is the weight of MnO2 = 1kg =1000g [given]
                                              F = Faradays constant
                                            t = time required
                                          Z =no.of electrons transfered
                                        M = mass of MnO₂ =86.9g/mol
​Substituting these values we get:

$$\begin{gathered} I=1000g\left[\frac{96500C/mol\times 2}{24\times 60\times 60\times 86.9}\right] \\ =25.70 A \end{gathered}$$

Now Current efficiency
           = $$\begin{gathered} \frac{25.7}{27}\times 100\% \\ =95.185\% \end{gathered}$$

Hence current efficiency = 95.185%
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