Solveeit Logo
Login

Question

Question: When \(PbO_{2}\)volt and \(H_{2}SO_{4}\) volt, which of the following is correct....

When PbO2PbO_{2}volt and H2SO4H_{2}SO_{4} volt, which of the following is correct.

A

PbO2PbO_{2} can be reduced by H2SO4H_{2}SO_{4}

B

Pb(s)+SO42−(aq)→PbSO4(s)+2e−Pb_{(s)} + S{O_{4}^{2 -}}_{(aq)} \rightarrow PbSO_{4(s)} + 2e^{-} can oxidise 2SO4(aq)2−→S2O8(aq)2−+2e−2SO_{4(aq)}^{2 -} \rightarrow S_{2}O_{8(aq)}^{2 -} + 2e^{-} into $2H_{2(g)} + 4O{H^{-}}{(aq)} \rightarrow $$$4H{2}O_{(l)} + 4e^{-}$$

C

H2SO4H_{2}SO_{4} can be reduced by H2SO4H_{2}SO_{4}

D

2Fe(s)+O2(g)+4H+(aq)→2F(aq)2++2H2O(l)2Fe_{(s)} + O_{2(g)} + 4{H^{+}}_{(aq)} \rightarrow 2F_{(aq)}^{2 +} + 2H_{2}O_{(l)} can reduce →\rightarrow ion

Answer

PbO2PbO_{2} can be reduced by H2SO4H_{2}SO_{4}

Explanation

Solution

Because H2H_{2} has greater reduction potential so it reduced the Ag+Ag^{+}.