The emf of a Daniell cell at 298 K is E1 Zn/ZnSO4(0.01 text M)||CuSO4 text(1 text.0 textM) ! !| ! ! text Cu When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M. The emf changed to E2 . What is the relation between E1 and E2 ?

Question

The emf of a Daniell cell at 298 K is E1 Zn/ZnSO4(0.01 text M)||CuSO4 text(1 text.0 textM) ! !| ! ! text Cu When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M. The emf changed to E2 . What is the relation between E1 and E2 ? (A)  E1=E2  (B)  E2=0≠ E1  (C)  E1>E2  (D)  E1<E2

Tardigrade Admin · Accepted Answer

E2=E1-(0.0591/n) log ([Zn2+]/[Cu2+]) E2=E1-(0.0591/2) log (1/0.01) E2=E1-0.0591 Hence, E1>E2

Q. The emf of a Daniell cell at 298 K is $E_{1}$ $Z n / Z n S O_{4} (0.01 M) ∣∣ C u S O_{4} (1 .0 M) ∣ C u$ When the concentration of $Z n S O_{4}$ is 1.0 M and that of $C u S O_{4}$ is 0.01 M. The emf changed to $E_{2}$ . What is the relation between $E_{1}$ and $E_{2}$ ?

$E_{1} = E_{2}$

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

$E_{1} > E_{2}$

$E_{1} < E_{2}$

none of these

$E_{2} = E_{1} - \frac{0.0591}{n} lo g \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ $E_{2} = E_{1} - \frac{0.0591}{2} lo g \frac{1}{0.01}$ $E_{2} = E_{1} - 0.0591$ Hence, $E_{1} > E_{2}$

Q. The emf of a Daniell cell at 298 K is E1​ Zn/ZnSO4​(0.01M)∣∣CuSO4​(1.0M) ∣Cu When the concentration of ZnSO4​ is 1.0 M and that of CuSO4​ is 0.01 M. The emf changed to E2​ . What is the relation between E1​ and E2​ ?

E1​=E2​

E2​=0=E1​

E1​>E2​

E1​<E2​