<em>K</em> = 5.0 × 10^25
<h2>Part (a). Calculate <em>E</em>° for the reaction
</h2>
<em>Step 1.</em> Write the equations for the two half-reactions
2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V
Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V
<em>Step 2.</em> Identify the cathode and the anode
The half-cell with the more negative <em>E</em>° (Zn) is the anode.
<em>Step 3.</em> Calculate <em>E</em>°
Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V
2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V
Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V
<em>E</em>° = +0.76 V
<h2>Part (b). Calculate <em>K</em> for the reaction
</h2>
The relation between <em>E</em>° and <em>K</em> is
<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K
</em>
where
<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)
<em>T</em> = the Kelvin temperature
<em>n</em> = the moles of electrons transferred
<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)
Then
0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>
0.76 = 0.012 85 ln<em>K</em>
ln<em>K</em> = 0.76/0.012 85 = 59.16
<em>K</em> =e^59.16 = 5.0 × 10^25
