Standard electrode potential for the cell with cell reaction Zn ( s )+ Cu 2+( aq ) arrow Zn 2+( aq )+ Cu ( s ) is 1.1 V. Calculate the standard gibbs energy change for the cell reaction. (Given F =96487 C mol -1 )

Question

Standard electrode potential for the cell with cell reaction Zn ( s )+ Cu 2+( aq ) arrow Zn 2+( aq )+ Cu ( s ) is 1.1 V. Calculate the standard gibbs energy change for the cell reaction. (Given F =96487 C mol -1 ) (A) -200.27 kJ mol -1 (B) -212.27 kJ mol -1 (C) -212.27 J mol -1 (D) -200.27 J mol -1

Tardigrade Admin · Accepted Answer

Δ G °=- nF ⋅ E text cell ° Δ G °=(-2 × 96487 × 1.1/1000) Δ G °=-212.27 kj mol -1

Q. Standard electrode potential for the cell with cell reaction
$Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)$ is $1.1 V$ .
Calculate the standard gibbs energy change for the cell reaction.
(Given $F = 96487 C m o l^{- 1}$ )

$- 200.27 k J m o l^{- 1}$

$- 212.27 k J m o l^{- 1}$

$- 212.27 J m o l^{- 1}$

$- 200.27 J m o l^{- 1}$

$Δ G^{\circ} = - n F \cdot E_{cell}^{\circ}$
$Δ G^{\circ} = \frac{- 2 \times 96487 \times 1.1}{1000}$
$Δ G^{\circ} = - 212.27 kj m o l^{- 1}$

Q. Standard electrode potential for the cell with cell reaction Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) is 1.1V. Calculate the standard gibbs energy change for the cell reaction. (Given F=96487Cmol−1 )

−200.27kJmol−1

−212.27kJmol−1

−212.27Jmol−1

−200.27Jmol−1