# Question 2bd1a

Apr 19, 2017

E_text(cell)^@ = "1.66 V"; ΔG^@ = "-480 kJ·mol"^"-1"; K = 2 × 10^84

#### Explanation:

Here's a useful diagram that shows the relation between the three quantities.

Calculate the standard cell potential

$\textcolor{w h i t e}{m m m m m m m m m m m m m m m m m m m m l l} {E}^{\circ} \text{/V}$
3/2×["2H"^"+""(aq)" + "2e"^"-" → "H"_2"(g)"]color(white)(mmmmmml)0.00
ul(1 ×["Al(s)" + "3e"^"-" → "Al"^"3+""(aq)"])color(white)(mmmmmm)ul("+1.66")
$\text{Al(s)" + "3H"^"+""(aq)" → "Al"^"3+""(aq)" + 3/2"H"_2"(g)"color(white)(mll)"+1.66}$

Calculate ΔG^@

ΔG^@ = "-"nFE_text(cell)^@ = "-3" × "96 485 J"·color(red)(cancel(color(black)("V")))^"-1""mol"^"-1" × 1.66 color(red)(cancel(color(black)("V"))) = "-480 000 J·mol"^"-1" = "-480 kJ·mol"^"-1"

Calculate $K$

${E}^{\circ} = \frac{R T}{n F} \ln K$

lnK = (nFE^@)/(RT) = (3 × "96 485" color(red)(cancel(color(black)("J·V"^"-1""mol"^"-1"))) × 1.66 color(red)(cancel(color(black)("V"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1")))× 298 color(red)(cancel(color(black)("K")))) = 194

K = e^194 = 2× 10^84#

That's a BIG equilibrium constant.