A galvanic cell has the overall reaction: zn(s) 2eu(no3)3(aq)  zn(no3)2(aq) 2eu(no3)2(aq) which is the half reaction occurring at the anode

Question

asked Apr 1, 2019 150k views

2 Answers

Tuespetre · Answer 1 · 2019-04-02T18:50:31+0000

Answer: The half reaction occurring at anode is
$Zn(s)\rightarrow Zn^(2+)(aq.)+2e^-$

Step-by-step explanation:

In a chemical cell, substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

$X\rightarrow X^(n+)+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

$X^(n+)+ne^-\rightarrow X$

For the given chemical cell reaction:

$Zn(s)+2Eu(NO_3)_3(aq.)\rightarrow Zn(NO_3)_2(aq.)+2Eu(NO_3)_2(aq.)$

Oxidation half reaction (anode):
$Zn(s)\rightarrow Zn^(2+)(aq.)+2e^-$

Reduction half reaction (cathode):
$Eu^(3+)(aq.)+e^-\rightarrow Eu^(2+)(aq.)$ ( × 2 )

Here, zinc is loosing electrons and is getting oxidized and europium is gaining electrons and loosing electrons.

Hence, the half reaction occurring at anode is
$Zn(s)\rightarrow Zn^(2+)(aq.)+2e^-$

Travel · Answer 2 · 2019-04-06T01:05:56+0000

The half reaction that occur at the anode is

Zn (s)→ Zn²⁺ (aq) + 2e⁻

Explanation

In galvanic cell oxidation take place at the the anode. That is there is lost of electrons at the anode.

Zn is therefore oxidized since it's oxidation move from oxidation state 0 to oxidation state 2+.