Question

Redox balancing

Answer 1

I have a student, she's been struggling with a specific redox reaction from her course book. I have personally not been able to figure out what's wrong here, any help will be appreciated:

Balance the follwing redox reaction

\[... S ^{2-}+ ...I_2 + ...OH ^{-} \rightarrow ...SO_4^{2-}+...I^-+...H_2 O\]

Answer 2

@Vocaloid

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
sulfur gets oxidized from an oxidation state of -2 to +6
iodine gets reduced from an oxidation state of 0 to -1

therefore your half equations are

\[S^{2-} \rightarrow SO_{4}^{2-}\]
\[I_{2} \rightarrow 2I^{-}\]

No hydrogen atoms are involved, so move to balancing oxygen
\[S^{2-} +4H_{2}O\rightarrow SO_{4}^{2-}\]
Re-balance hydrogen

\[S^{2-} +4H_{2}O\rightarrow SO_{4}^{2-} +8H^{+}\]
Balance charges

\[S^{2-} +4H_{2}O\rightarrow SO_{4}^{2-}+8e^{-}+8H^{+}\]
\[I_{2} + 2e^{-} \rightarrow 2I^{-}\]

Answer 4

balance charges so equations can be added

\[S^{2-} +4H_{2}O\rightarrow SO_{4}^{2-}+8e^{-}+8H^{+}\]
\[4I_{2} + 8e^{-} \rightarrow 8I^{-}\]

Add equations:

\[S^{2-} +4H_{2}O+4I_{2} \rightarrow SO_{4}^{2-}+8H^{+}+8I^{-}\]

Doesn't specify whether it's acidic or basic environment but just in case you have a basic environment: add +8OH- to each side

\[S^{2-} +4H_{2}O+4I_{2}+8OH^{-} \rightarrow SO_{4}^{2-}+8H^{+}+8OH^{-}+8I^{-}\]
\[S^{2-} +4H_{2}O+4I_{2}+8OH^{-} \rightarrow SO_{4}^{2-}+8H_{2}O+8I^{-}\]
\[S^{2-} +4I_{2}+8OH^{-} \rightarrow SO_{4}^{2-}+4H_{2}O+8I^{-}\]