Daniell cell is shown as

Question

Tardigrade Admin · Accepted Answer

The overall reaction of Daniell cell is Zn (s)+ Cu 2+ (a q) arrow Cu (s)+ Zn 2+(aq) So, its cell representation will be as follows: Zn (s)| Zn 2+ (a q) | Cu 2+ (a q)| Cu (s)

Q. Daniell cell is shown as

$Z n_{(s)} ∣ Z n^{2 +}_{(a q)} ∥ C u_{(s)} ∣ C u_{(a q)}^{2 +}$

$Z n_{(a q)}^{2 +} ∣ ∣ Z n_{(s)} ∥ C u_{(s)} ∣ ∣ C u^{2 +} (a q)$

$Z n_{(s)} ∣ ∣ Z n^{2 +}_{(a q)} ∥ C u^{2 +} (a q) ∣ ∣ C u_{(s)}$

$Z n_{(s)} ∣ ∣ C u^{2 +}_{(a q)} ∥ Z n^{2 +}_{(a q)} ∣ ∣ C u_{(s)}$

The overall reaction of Daniell cell is

$Z n_{(s)} + C u^{2 +}_{(a q)} \to C u_{(s)} + Z n_{(a q)}^{2 +}$

So, its cell representation will be as follows:

$Z n_{(s)} ∣ ∣ Z n^{2 +}_{(a q)} ∥ C u^{2 +}_{(a q)} ∣ ∣ C u_{(s)}$

Q. Daniell cell is shown as

Zn(s)​∣Zn2+(aq)​∥Cu(s)​∣Cu(aq)2+​

Zn(aq)2+​∣∣​Zn(s)​∥Cu(s)​∣∣​Cu2+(aq)

Zn(s)​∣∣​Zn2+(aq)​∥Cu2+(aq)∣∣​Cu(s)​

Zn(s)​∣∣​Cu2+(aq)​∥Zn2+(aq)​∣∣​Cu(s)​

The overall reaction of Daniell cell is Zn(s)​+Cu2+(aq)​→Cu(s)​+Zn(aq)2+​ So, its cell representation will be as follows: Zn(s)​∣∣​Zn2+(aq)​∥Cu2+(aq)​∣∣​Cu(s)​

$Z n_{(s)} ∣ Z n^{2 +}_{(a q)} ∥ C u_{(s)} ∣ C u_{(a q)}^{2 +}$

$Z n_{(a q)}^{2 +} ∣ ∣ Z n_{(s)} ∥ C u_{(s)} ∣ ∣ C u^{2 +} (a q)$

$Z n_{(s)} ∣ ∣ Z n^{2 +}_{(a q)} ∥ C u^{2 +} (a q) ∣ ∣ C u_{(s)}$

$Z n_{(s)} ∣ ∣ C u^{2 +}_{(a q)} ∥ Z n^{2 +}_{(a q)} ∣ ∣ C u_{(s)}$

The overall reaction of Daniell cell is

$Z n_{(s)} + C u^{2 +}_{(a q)} \to C u_{(s)} + Z n_{(a q)}^{2 +}$

So, its cell representation will be as follows:

$Z n_{(s)} ∣ ∣ Z n^{2 +}_{(a q)} ∥ C u^{2 +}_{(a q)} ∣ ∣ C u_{(s)}$