Question

Calculate the EMF generated by a cell given by the reaction below when [Al³⁺] = 4.0 x 10^-3M and [I^-] = 0.010 M

2 Al (s)     +     3 I₂ (s)        →       2 Al³⁺ (aq)    +     6 I^- (aq)

Answer 1

Answer 2

Calculating EMF of the Given Cell

Given Reaction:

$$2\text{Al (s)}+3{\text{I}}_2\text{(s)}\to 2{\text{Al}}^{3+}\text{(aq)}+6{\text{I}}^{-}\text{(aq)}$$

• Concentrations:

• $[{\text{Al}}^{3+}]=4.0\times 10^{-3}\text{M}$

• $[{\text{I}}^{-}]=0.010\text{M}$

Step 1: Identify Half-Reactions

1. Oxidation (Anode):

   $$\text{Al (s)}\to {\text{Al}}^{3+}\text{(aq)}+3e^{-}{E}_{\text{anode}}^{\circ }=-1.66\text{V}$$

2. Reduction (Cathode):

   $${\text{I}}_2\text{(s)}+2e^{-}\to 2{\text{I}}^{-}\text{(aq)}{E}_{\text{cathode}}^{\circ }=+0.54\text{V}$$

Step 2: Calculate Standard EMF ($E_{\text{cell}}^{\circ }$)

$$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=0.54\text{V}-(-1.66\text{V})=2.20\text{V}$$

Step 3: Apply the Nernst Equation

$$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{n}\log Q$$

• Reaction Quotient ($Q$):

  $$Q=\frac{[{\text{Al}}^{3+}{]}^2[{\text{I}}^{-}{]}^6}{1}=(4.0\times 10^{-3}{)}^2(0.010{)}^6=1.6\times 10^{-20}$$

• Electrons Transferred ($n$):
  $n=6$ (balanced equation shows 6 $e^{-}$).

$$E_{\text{cell}}=2.20-\frac{0.0591}{6}\log (1.6\times 10^{-20})$$$$E_{\text{cell}}=2.20-0.00985\times (-19.8)\approx 2.20+0.195=2.395\text{V}$$

Final Answer:

$$\boxed{{\displaystyle 2.40\text{V}}}\text{(Rounded to 2 decimal places)}$$