Balance the reaction between Ag2O and I^- to form I2 and Ag in basic solution. When you have...

The oxidation half-reaction is shown below:

{eq}I^{-}\rightarrow I_{2} {/eq}

Balancing {eq}I {/eq}, we get: {eq}2I^{-}\rightarrow I_{2} {/eq}

Balancing charge, we get: {eq}2I^{-}\rightarrow I_{2}+2e^- {/eq}

So, the balanced oxidation half-reaction is:

{eq}2I^{-}\rightarrow I_{2}+2e^- {/eq} ............... (1)

The reduction half-reaction is shown below:

{eq}Ag_{2}O\rightarrow Ag {/eq}

Balancing {eq}Ag {/eq}, we get: {eq}Ag_{2}O\rightarrow 2Ag {/eq}

Balancing {eq}O {/eq} and {eq}H {/eq} in the basic medium, we get: {eq}Ag_{2}O+H_2O\rightarrow 2Ag+2OH^- {/eq}

Balancing charge, we get: {eq}Ag_{2}O+H_2O+2e^-\rightarrow 2Ag+2OH^- {/eq}

The balanced reduction half-reaction is:

{eq}Ag_{2}O+H_2O+2e^-\rightarrow 2Ag+2OH^- {/eq} ................. (2)

{eq}Equation-(1)+Equation-(2): {/eq}

{eq}Ag_{2}O+2I^{-}+H_2O\rightarrow 2Ag+I_2+2OH^- {/eq}

So, the balanced equation ("1" is explicitly shown) is shown below:

{eq}1Ag_{2}O+2I^{-}+1H_2O\rightarrow 2Ag+1I_2+2OH^- {/eq}

Water appears in the balanced equation as a reactant with a coefficient of 1.

Two electrons are transferred in the given reaction.