Which is stronger oxidizing agent
(i). `K_(2)Cr_(2)O_(7)` in solution in which
`[Cr_(2)O_(7)^(2-)]=0.1M,[Cr^(3+)]=10^(-2)M` and `[H^(+)]=10^(-1)M`
(ii). `KMnO_(4)` in a solution in which
`[MnO_(4)^(-)]=10^(-1)M,[Mn^(2+)]=10^(-2)M,[H^(+)]=10^(-2)M`
`E_(Cr_(2)O_(7)^(2-)//Cr^(+3))^(0)=1.33V,E_(MnO_(4)^(-)//Mn^(+2))^(0)=1.51V`

Explanation

To determine which of the two compounds, \( K_2Cr_2O_7 \) or \( KMnO_4 \), is the stronger oxidizing agent, we will calculate the reduction potential for each compound using the Nernst equation.

### Step 1: Write the half-reactions and standard reduction potentials

1. **For \( K_2Cr_2O_7 \):**
The half-reaction is:
\[
Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O
...