Question

Calculate δg o for the reaction between al(s) and cd2+(aq). enter your answer in scientific notation.

Answer 1

The reaction between Al(s) and Cd²⁺(aq) is:
Al(s) + Cd²⁺(aq) → Al³⁺(aq) + Cd(s)

The half reactions are:
Al³⁺(aq) + 3 e⁻ → Al(s)           E⁰ = -1.66 V
Cd²⁺(aq) + 2 e⁻ → Cd(s)        E⁰ = -0.40 V

Reverse the first equation and add to the second one
2 Al(s) + 3 Cd²⁺(aq) → 2 Al³⁺(aq) + 3 Cd(s)

So, the overall cell potential is:
E⁰cell = 1.66 - 0.40 = 1.26 V

Free energy of the reaction is:
ΔG⁰ = -n F E⁰cell
        = - 6 mol e⁻ * (96485 C / mol e⁻) * (1.26 V)
        = - 7.29 x 10⁵ J

Answer 2

The thermodynamic potential is given by the Gibbs free energy at constant pressure and temperature. The free energy of the reaction is $-7.29 \times 10 ^{5}\;\rm J$.

What is Gibbs free energy?

Gibbs free energy is given as the difference of the change in the enthalpy with the product of the temperature and change in the entropy.

The chemical reaction Aluminium and cadmium ion can be shown as,

$\rm Al(s) + Cd^{2+}(aq) \rightarrow Al^{3+}(aq) + Cd(s)$

The half-reactions of the chemical equation are:

$\begin{aligned}\rm Al^{3+}(aq) + 3 e^{-} \rightarrow Al(s)\; ,E^{\circ} &= -1.66 \;\rm V\\\\\rm Cd^{2+}(aq) + 2 e^{-} \rightarrow Cd(s)\; ,E^{\circ} &= -0.40 \;\rm V\end{aligned}$

The overall reaction is shown as,

$\rm 2 Al(s) + 3 Cd^{2+}(aq) \rightarrow 2 Al^{3+}(aq) + 3 Cd(s)$

From the overall reaction the overall cell potential will be:

$\begin{aligned}\rm E^{\circ} cell &= 1.66 - 0.40 \\\\&= 1.26 \;\rm V \end{aligned}$

Free energy for the reaction is calculated as:

$\begin{aligned}\rm \Delta G^{\circ} &= \rm -n F E^{\circ} cell\\\\& = - 6 \;\rm mol e^{-} \times (96485 C / mol e^{-}) \times (1.26 V)\\\\& = - 7.29 \times 10^{5}\;\rm J\end{aligned}$

Therefore, Gibbs free energy is $-7.29 \times 10 ^{5}\;\rm J.$

Learn more about Gibbs free energy here:

https://brainly.com/question/15722525