The emf of a Daniel cell at 298 K is E1. qquadThe cell is Zn | ZnSO4 (0.01M)|| CuSO4 (1M) | CU When the concentration of ZnSO4 is changed to 1M and that of CuSO4 to 0.01M, the emf changes to E2.The relationship between E1 and E2 will be

Question

The emf of a Daniel cell at 298 K is E1. qquadThe cell is Zn | ZnSO4 (0.01M)|| CuSO4 (1M) | CU When the concentration of ZnSO4 is changed to 1M and that of CuSO4 to 0.01M, the emf changes to E2.The relationship between E1 and E2 will be (A) E1-E2 =0 (B) E1 < E2 (C) E1>E2 (D) E1=102 E2

Tardigrade Admin · Accepted Answer

Zn+CuSO4 leftharpoons ZnSO4 +cu n = 2 Q=(Zn2+/cu2+) E1-E2 =-(0.059/2) (log(0.01/1)log(1/0.01)) =-(0.059/2)(log(1/100)-log100)=(0.059/2) (-log100 -log 100)=+(0.059/2)×2 so, E1 >E2

Q. The emf of a Daniel cell at 298 K is $E_{1} .$ The cell is
$Z n ∣ Z n S O_{4} (0.01 M) ∣∣ C u S O_{4} (1 M) ∣ C U$
When the concentration of $Z n S O_{4}$ is changed to 1M and that of $C u S O_{4}$ to 0.01M, the emf changes to E2.The relationship between $E_{1} an d E_{2}$ will be

$E_{1} - E_{2} = 0$

$E_{1} < E_{2}$

$E_{1} > E_{2}$

$E_{1} = 1 0^{2} E_{2}$

Q. The emf of a Daniel cell at 298 K is E1​.The cell is Zn∣ZnSO4​(0.01M)∣∣CuSO4​(1M)∣CU When the concentration of ZnSO4​ is changed to 1M and that of CuSO4​ to 0.01M, the emf changes to E2.The relationship between E1​andE2​ will be

E1​−E2​=0

E1​<E2​

E1​>E2​

E1​=102E2​

Zn+CuSO4​⇌ZnSO4​+cu n = 2 Q=cu2+Zn2+​ E1​−E2​=−20.059​(log10.01​log0.011​)=−20.059​(log1001​−log100)=20.059​(−log100−log100)=+20.059​×2 so,E1​>E2​

Q. The emf of a Daniel cell at 298 K is $E_{1} .$ The cell is
$Z n ∣ Z n S O_{4} (0.01 M) ∣∣ C u S O_{4} (1 M) ∣ C U$
When the concentration of $Z n S O_{4}$ is changed to 1M and that of $C u S O_{4}$ to 0.01M, the emf changes to E2.The relationship between $E_{1} an d E_{2}$ will be

$E_{1} - E_{2} = 0$

$E_{1} < E_{2}$

$E_{1} > E_{2}$

$E_{1} = 1 0^{2} E_{2}$