Consider the following redox reaction:
`2lrCl_(6)^(3-)+3HCOOHto2lr+3CO_(2)+12Cl^(-)+6H^(+)`
Given: `CO_(2)+2H_(3)O^(+)+2etoHCOOH+2H_(2)O" "E^(@)=-0.20V`
`lrCl_(6)^(3-)+3etolr+6Cl^(-)" "E^(@)=0.77V`
(a). Determine standard state emf of cell.
(b). Is this reaction thermodynamically spontaneous as written? Briefly explain.

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.

The given redox reaction is:
\[ 2 \text{IrCl}_6^{3-} + 3 \text{HCOOH} \rightarrow 2 \text{Ir} + 3 \text{CO}_2 + 12 \text{Cl}^- + 6 \text{H}^+ \]

From the information provided, we have the following half-reactions and their standard electrode potentials:

1. **Reduction half-reaction**:
\[ \text{IrCl}_6^{3-} + 3e^- \rightarrow \text{Ir} + 6 \text{Cl}^- \]
\[ E^\circ_{\text{reduction}} = 0.77 \, \text{V} \]

2. **Oxidation half-reaction**:
\[ \text{CO}_2 + 2 \text{H}_2O + 2e^- \rightarrow \text{HCOOH} + 2 \text{H}^+ \]
\[ E^\circ_{\text{oxidation}} = -0.20 \, \text{V} \]

### Step 2: Calculate the standard state emf of the cell.

The standard cell potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.77 \, \text{V} - (-0.20 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.77 \, \text{V} + 0.20 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 0.97 \, \text{V} \]

### Step 3: Determine if the reaction is thermodynamically spontaneous.

To determine if the reaction is thermodynamically spontaneous, we can use the relationship between the standard cell potential and the Gibbs free energy change (\( \Delta G^\circ \)):
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( n \) = number of moles of electrons transferred in the balanced equation.
- \( F \) = Faraday's constant (approximately \( 96485 \, \text{C/mol} \)).

Since \( E^\circ_{\text{cell}} = 0.97 \, \text{V} \) is positive, it indicates that \( \Delta G^\circ \) is negative, which means the reaction is spontaneous.

### Final Answers:
(a) The standard state emf of the cell is \( 0.97 \, \text{V} \).

(b) Yes, the reaction is thermodynamically spontaneous as written because \( \Delta G^\circ < 0 \).