Galvanic cell

$E_{cell}=\phi_{R} - \phi_{L}$

$Cu(s)+2Ag^+ (aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$

Cathode (RHE) reduction $2Ag^+(aq) + 2e \rightarrow 2Ag(s) \Rightarrow Ag^+(aq)/Ag(s)$

Anode (LHE) Oxidation $Cu(s) \rightarrow Cu^{2+}(aq)+2e \Rightarrow Cu/Cu^{2+}(aq)$

$Cu | Cu^{2+}(aq)$ | | $Ag^+ (aq) | Ag(s)$

$Cu | Cu^{2+}(aq) \rightarrow LHE \rightarrow Oxidation$

$Ag^+ (aq) | Ag(s) \rightarrow RHE \rightarrow Reduction$

$E_{cell} = \phi_R - \phi_L = \phi_{Ag^+/Ag} - \phi_{Cu^{2+}/Cu}$

$Zn(S)|ZnSO_4(C=1)$ | | $CuSO_4(C=1)|Cu$

Measurement of half-cell potential

Standard hydrogen electrode

$Pt(s) | H_2(g)(P=1bar)|H^+(aq,1M)$

$\phi = 0$ at all T

$H^+(aq) + e \rightarrow \frac{1}{2} H_2(g,1bar)$

$E = \phi_R - \phi_L$

$E=\phi_{Ag^+/Ag}-\phi_{Cu^{+2}/Cu}$

$E_{cell} = \phi_R - \phi_L = \phi_{Ag^+/Ag}-\phi_{Cu^{2+}/Cu}$

SHE $\quad \quad$ || $\quad$ 2nd Half cell

$\quad \darr \quad \quad\quad \quad \quad \quad \darr$

Anode $\qquad \quad \quad$ Cathode

LHE $\qquad \quad \quad\quad$ RHE

$LHE \rightarrow H^+(aq) + e \rightarrow \frac{1}{2}H_2 (g,1bar)$

$RHE \rightarrow$ reduction potential of the RH half cell

Concentration of the redox active substances/electrolyte in set at unity

Cell potential /Cell EMF = standard electrode potential or, standard reduction potential of the half cell

$E^{\phi} = \phi_R^{\phi} - \phi_L ^{\phi}$

$E^{\phi} = \phi_R ^{\phi}$

If ${\phi_L ^{\phi} }$ will be zero

$Cu^{2+}(aq, 1M)|Cu$

$SHE || Cu^{2+} (aq,1M)|Cu(s)$ at 298K

$Pt(s) | H_2(g, 1bar) H^+(aq, 1M)$ | | $Cu^{2+} (aq, 1M)|Cu(s)$

SHE/LHE/oxidation

$E_{cell} ^{\phi} = \phi_R = 0.34V$

$Cu^{2+} (aq,1M) + 2e^- \rightarrow Cu^0 (S)$

$\Delta G^0=-nFE^0$

$\Delta G^0 < 0$

$Pt(S) |H_2(g, 1bar) (H^+ (aq, 1M) || Z^{2+} (aq, 1M) | Zn(s) \quad \quad \quad 298K$

$LHE \quad \quad \quad \quad \quad \quad \quad \quad RHE$

Anode (Oxidation take place)

$E^o=-0.76V$

Cathod (Reduction)

$Zn^{2+} + 2e \rarr Zn^0$

$\phi_{Zn^{2+} /Zn}=-0.76v$

$\Delta G^0 = -nFE^0\Rightarrow 0$

$Zn(s)|(ZnSO_4 (C=1) || CuSO_4 (c=1) | Cu$

$-0.761 \qquad \qquad \qquad \qquad 0.34v$

$E_{cell} = \phi_{Cu^{2+}/Cu} - \phi_{Zn^{2+}/Zn}$

$=0.34 -(-0.76)V = 1.1V$

$Cu|Cu^{2+}(C = 1)$ Reduction potential $= 0.34V$

Reduction potential $\rightarrow Cu^{2+}\rarr Cu^o$

$Zn|Zn^{2+}(C=1)$ $\quad \quad -0.76V$

$Zn^{2+} + 2e\rarr Zn^o \quad \Delta G^0 > 0$

$H_2$ - electrode : SHE

$Br_2$ - electrode : $Pt(S)|Br_2(aq)|Br^-(aq)$

$\frac {1}{2}Br_2(aq) +e \rarr Br^-(aq)$

$Ag(S) | Ag^+ (C =…)|$

$Fe^{3+} +e \rarr Fe^{2+}$

AgCl $+e^{-} \rightarrow Ag^o+Cl^-$

$Ag|AgCl(S),Cl^-$

$C=1 \quad \Delta G -nFE$

NERNST Equation

$\phi_{M^{n+}/M} = \phi_\frac {M^{n+}}{M}$ - $\frac {RT}{nF}ln \frac{[M]}{[M^{n+}]}$

$M^{n+} + ne^{-} \rarr M$

$F = \sim 96500$ C/mole & $R = 8.314 jK^{-1}mole^{-1}$

$\phi_{M^{n+}/M} = \phi_{M^{n+}/M}^o - \frac{RT}{nF}ln \frac{1}{[M^{n+}]}$

Cathode RHE Reduction

$\phi_{Cu^{2+}/Cu} = \phi^0_{Cu^{2+}/{Cu}} - \frac {RT}{2F} ln \frac {1}{[Cu^{2+}aq]}$

Anode LHE Oxidation

$\phi_{Zn^{2+} /{Zn} } =\phi_{Zn^{2+} /{Zn}}- \frac {RT}{2F}ln \frac {1}{Zn^{2+}aq}$

$=\phi_{Cu^{2+} /{Cu}} - \phi_{Zn^{2+} /{Zn}}$

$E_{cell}$ = $\phi_R-\phi_L$ (IUPAC, Reduction pot)

$= \phi_{Cu^{2+} /Cu} -\phi_{Zn^{2+} /{Zn}}$

$= \phi_{Cu^{2+} /Cu} ^0 -\phi_{Zn^{2+} /{Zn}} ^0 -\frac {RT}{2f}ln \frac{1}{Zn^{2+} (aq)} + \frac{RT}{2F} ln \frac{1} {Zn^{2+} (aq)}$

$E_{cell}= E^0_{cell} - \frac {RT}{2F}ln \frac {[Zn^{2+}(aq)]}{[Cu^{2+}(aq)]}$

$Zn + CuSO_4\rarr ZnSO_4 + Cu$

R,F,T = 298K

$E_{cell}=E^o_{cell}-\frac{0.059}{2} log \frac{[Zn^{2+}(aq)]}{[Cu^{2+}(aq)]}$