How to separate this redox reaction into two half reactions?

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Dhamnekar Winod
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How to separate this redox reaction into two half reactions?

How to separate this redox reaction into two half reactions?

$$HCHO(l) + 2[Ag(NH_3)_2]^+(aq) + 3OH^-(aq) \rightarrow 2Ag(s) + HCOO^-(aq) + 4NH_3(aq) + 2H_2O(l)$$
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
ChenBeier
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Re: How to separate this redox reaction into two half reactions?

HCHO + 3 OH- => HCOO- + 2 H2O + 2 e- Oxidation

[Ag(NH3)2]+ + e- => Ag + 2 NH3 Reduction
Dhamnekar Winod
Sr. Member
Posts: 35
Joined: Sat Nov 21, 2020 10:14 am
Location: Mumbai[Bombay],Maharashtra State,India

Re: How to separate this redox reaction into two half reactions?

You have given oxidation and reduction reactions in your reply. But in other chemistry notes, these reactions are given as below:

$$H_2O(l) + HCHO(l) \rightarrow HCOO^-(aq) +3H^+ +2e^-\Rightarrow$$ OXIDATION

$$[Ag(NH_3)_2]^+ (aq) +e^- \rightarrow Ag(s) + 2NH_3(aq) + 2H_2O(l) \Rightarrow$$ REDUCTION

What is your explanation for these half reactions?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
ChenBeier
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Re: How to separate this redox reaction into two half reactions?

It depends which pH you have. If its acidic environment then H+ is used, in alcaline condition OH-.

But the right one is the alcaline one in this case. The Diamminsilver is not stable in acidic condition.
Dhamnekar Winod
Sr. Member
Posts: 35
Joined: Sat Nov 21, 2020 10:14 am
Location: Mumbai[Bombay],Maharashtra State,India

Re: How to separate this redox reaction into two half reactions?

We get the following redox reactions by adding the two half reactions $$H_2O (l) + HCHO(l) \rightarrow HCOO^-(aq) + 3H^+ + 2e^- \tag{Oxidation}$$ and $$[Ag(NH_3)_2]^+(aq) + e^- \rightarrow Ag(s) + 2NH_3(aq) + 2H_2O(l) \times 2 \tag{Reduction}$$

$$= HCHO(l) + 2[Ag(NH_3)_2]^+(aq) \rightarrow HCOO^-(aq) +3H^+ + 2Ag(s) +4NH_3(aq) + 3H_2O(l)$$

which is different from the original redox equation namely $$HCHO(l) + 2[Ag(NH_3)_2]^+ (aq) + 3OH^- (aq) \rightarrow 2Ag(s) + HCOO^-(aq) +4NH_3(aq) + 2H_2O(l)$$

Where is the mistake? where are we wrong? Why did such discrepancy occur?

Would any chemistry expert answer these questions?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
ChenBeier
Distinguished Member
Posts: 537
Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany

Re: How to separate this redox reaction into two half reactions?

Did you read my answer above. You are wrong because [Ag(NH3)2] + is not existing in acidic conditions.

Mathematically you can write both equations, but only one will happen in practise.