In the electrochemical cell Zn|.ZnSO4(.0.01M.)||CuSO4(.1.0M.)|.Cu, the emf of this Daniel cell is E1 . When the concentration of ZnSO4 is changed to 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2 . From the followings, which one is the relationship between E1 and E2 ? (Given, (R T/F)=0.059 )

Question

In the electrochemical cell Zn|.ZnSO4(.0.01M.)||CuSO4(.1.0M.)|.Cu, the emf of this Daniel cell is E1 . When the concentration of ZnSO4 is changed to 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2 . From the followings, which one is the relationship between E1 and E2 ? (Given, (R T/F)=0.059 ) (A) E1 < E2 (B) E1>E2 (C) E2=0≠ E1 (D) E1=E2

Tardigrade Admin · Accepted Answer

E = E 0-(0.059/2) log 10 ([ Zn ]2+/[ Cu ]2+) E 1= E °-(0.059/2) log 10 ((0.01)/(1)2) E 2= E °-(0.059/2) log 10 ((1)/(0.01)2) Hence E 1> E 2

$E_{1} < E_{2}$

$E_{1} > E_{2}$

$E_{2} = 0 \neq = E_{1}$

$E_{1} = E_{2}$

$E = E^{0} - \frac{0.059}{2} lo g_{10} \frac{[ Z n ] ^{2 +}}{[ C u ] ^{2 +}}$

$E_{1} = E^{\circ} - \frac{0.059}{2} lo g_{10} \frac{( 0.01 )}{( 1 ) ^{2}}$

$E_{2} = E^{\circ} - \frac{0.059}{2} lo g_{10} \frac{( 1 )}{( 0.01 ) ^{2}}$

Hence $E_{1} > E_{2}$

E1​<E2​

E1​>E2​

E2​=0=E1​

E1​=E2​

E=E0−20.059​log10​[Cu]2+[Zn]2+​ E1​=E∘−20.059​log10​(1)2(0.01)​ E2​=E∘−20.059​log10​(0.01)2(1)​ Hence E1​>E2​

$E_{1} < E_{2}$

$E_{1} > E_{2}$

$E_{2} = 0 \neq = E_{1}$

$E_{1} = E_{2}$

$E = E^{0} - \frac{0.059}{2} lo g_{10} \frac{[ Z n ] ^{2 +}}{[ C u ] ^{2 +}}$

$E_{1} = E^{\circ} - \frac{0.059}{2} lo g_{10} \frac{( 0.01 )}{( 1 ) ^{2}}$

$E_{2} = E^{\circ} - \frac{0.059}{2} lo g_{10} \frac{( 1 )}{( 0.01 ) ^{2}}$

Hence $E_{1} > E_{2}$