In the cell, $Z n | Z n^{2 +} | | C u^{2 +} | C u$ , the negative terminal is:

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C u^{2 +}

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Z n^{2 +}

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Solution

The correct option is A Zn
As per the cell representation, the oxidation is written in the left part and the reduction is written to the right. Here, the left part i.e the oxidation occurs at anode. As in oxidation, the electrons are left behind on the electrode, hence the anode is the negative terminal.

Hence, here $Z n$ is the negative terminal.

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Similar questions

The cell representation of the given reaction is:
Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s)

For cell reaction, $Z n + C u^{2 +} \to Z n^{2 +} + C u,$ cell representation is

E_{1}, E_{2}, E_{3}

Z n | Z n_{1 M}^{2 +} | | C u_{0.1 M}^{2 +} | C u

Z n | Z n_{1 M}^{2 +} | | C u_{1 M}^{2 +} | C u

Z n | Z n_{0.1 M}^{2 +} | | C u_{1 M}^{2 +} | C u

Below given are emf of three galvanic cells, represented by $E_{1}$ , $E_{2}$ and $E_{3}$ . Which of the following is true?

I. $Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u$

II. $Z n | Z n^{2 +} (0.1 M) | | C u^{2 +} (1 M) | C u$

III. $Z n | Z n^{2 +} (1 M) | | C u^{2 +} (0.1 M) | C u$

E_{1}, E_{2}

E_{3}

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u

Z n | Z n^{2 +} (0.1 M) | | C u^{2 +} (1 M) | C u

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (0.1 M) | C u

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