Question

Balance each redox reaction occurring in acidic aqueous solution. Use the half-reaction method. Part A IO3−(aq)+SO2(g)→I2(s)+SO42−(aq)IO3−(aq)+SO2(g)→I2(s)+SO42−(aq) Express your answer as a chemical equation. Identify all of the phases in your answer. nothing SubmitRequest Answer Part B Sn4+(aq)+H2(g)→Sn2+(aq)+H+(aq)

Answer 1

Answer:

A. $2(I^{+5}O^{-2}_3)^-(aq)+5S^{+4}O^{-2}_2(g)+4H_2O\rightarrow I_2^0(s)+5(S^{+6}O^{-2}_4)^{2-}(aq)+8H^+$

B. $Sn^{4+}(aq)+H_2^0(g)\rightarrow Sn^{2+}(aq)+2H^+(aq)$

Explanation:

Hello,

A.) In this case, the reaction with each oxidation state is:

$(I^{+5}O^{-2}_3)^-(aq)+S^{+4}O^{-2}_2(g)\rightarrow I_2^0(s)+(S^{+4}O^{-2}_4)^{2-}(aq)$

Thus, the half reactions turn out:

$2(I^{+5}O^{-2}_3)^-(aq)+12H^++10e^-\rightarrow I_2^0(s)+6H_2O\\S^{+4}O^{-2}_2(g)+2H_2O\rightarrow (S^{+6}O^{-2}_4)^{2-}+2e^-(aq)+4H^+$

And the result:

$4(I^{+5}O^{-2}_3)^-(aq)+24H^++20e^-+10S^{+4}O^{-2}_2(g)+20H_2O\rightarrow 2I_2^0(s)+12H_2O+10(S^{+6}O^{-2}_4)^{2-}+20e^-(aq)+40H^+\\\\4(I^{+5}O^{-2}_3)^-(aq)+10S^{+4}O^{-2}_2(g)+8H_2O\rightarrow 2I_2^0(s)+10(S^{+6}O^{-2}_4)^{2-}(aq)+16H^+$

Simplifying:

$2(I^{+5}O^{-2}_3)^-(aq)+5S^{+4}O^{-2}_2(g)+4H_2O\rightarrow I_2^0(s)+5(S^{+6}O^{-2}_4)^{2-}(aq)+8H^+$

B.) In this case, the half reactions are:

$Sn^{4+}(aq)+2e^-\rightarrow Sn^{2+}(aq)\\H_2^0(g)\rightarrow 2H^++2e^-(aq)$

And the result:

$2Sn^{4+}(aq)+2H_2^0(g)+4e^-\rightarrow 2Sn^{2+}(aq)+4H^++4e^-(aq)\\Sn^{4+}(aq)+H_2^0(g)\rightarrow Sn^{2+}(aq)+2H^+(aq)$

Best regards.