07 question (1 point)\nthe value of $e^\\circ_{cell}$ for the following reaction is 0.500 v.\n$2mn^{3 +…

07 question (1 point)\nthe value of $e^\\circ_{cell}$ for the following reaction is 0.500 v.\n$2mn^{3 + }+2h_2o\\longrightarrow mn^{2 + }+mno_2 + 4h^{+}$\n2nd attempt\nwhat is the value of $\\delta g^\\circ_{cell}$ for this reaction? -96.5 kj\n1st attempt

07 question (1 point)\nthe value of $e^\\circ_{cell}$ for the following reaction is 0.500 v.\n$2mn^{3 + }+2h_2o\\longrightarrow mn^{2 + }+mno_2 + 4h^{+}$\n2nd attempt\nwhat is the value of $\\delta g^\\circ_{cell}$ for this reaction? -96.5 kj\n1st attempt

Answer

Explanation:

Step1: Identify the formula

The relationship between $\Delta G^{\circ}{cell}$ and $E^{\circ}{cell}$ is $\Delta G^{\circ}{cell}=-nFE^{\circ}{cell}$, where $n$ is the number of moles of electrons transferred, $F = 96485\ C/mol$ (Faraday's constant), and $E^{\circ}_{cell}$ is the standard - cell potential.

Step2: Determine the number of moles of electrons transferred ($n$)

For the reaction $2Mn^{3 +}+2H_2O\rightarrow Mn^{2+}+MnO_2 + 4H^{+}$, the oxidation state of $Mn$ changes from + 3 to + 2 (reduction) and from + 3 to + 4 (oxidation). The number of moles of electrons transferred $n = 2$.

Step3: Calculate $\Delta G^{\circ}_{cell}$

Substitute $n = 2$, $F=96485\ C/mol$, and $E^{\circ}{cell}=0.500\ V$ into the formula $\Delta G^{\circ}{cell}=-nFE^{\circ}{cell}$. $\Delta G^{\circ}{cell}=-2\times96485\ C/mol\times0.500\ V$. $\Delta G^{\circ}_{cell}=-96485\ J/mol=-96.5\ kJ/mol$

Answer:

$-96.5$