Write a balanced overall reaction from these unbalanced half-reactions.
sn-> sn2+ ag+-> ag balanced overall equation i got sn+2ag+-> sn+2ag, but its wrong.

The balanced overall equation for the two half reactions is:

We'll begin by balancing the various half reactions. This can be obtained as follow:

Sn → Sn²⁺ + 2e¯ (1)

Ag⁺ + e¯ → Ag (2)

Multiply equation (2) by 2 to balance the electron(s) on both equation

Sn → Sn²⁺ + 2e¯ (1)

2Ag⁺ + 2e¯ → 2Ag (2)

Combine equation 1 and 2 with the elimination of 2e¯ to obtain the overall reaction as shown below:

Sn + 2Ag⁺ + 2e¯ → Sn²⁺ + 2Ag + 2e¯

Sn + 2Ag⁺ → Sn²⁺ + 2Ag

Therefore, the balanced overall equation for the two half reactions is:

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Sn + 2Ag⁺ → Sn²⁺ + 2 Ag

Explanation:

1) The half-reactions given are:

Sn → Sn²⁺

Ag⁺ → Ag

2) The first step that you should perform is to complete each half reaction with electrons to balance the charges. This is how:

Sn → Sn²⁺ + 2e⁻

Ag⁺ + 1e⁻ → Ag

3) The second step is to multiply the half-reactions by the appropiate coefficients to make the electrons on both reactions equal.

So, multiply the second reaction by 2:

Sn → Sn²⁺ + 2e⁻

2Ag⁺ + 2e⁻ → 2Ag

4) The third step is to add up the two equations:

Sn + 2Ag⁺ + 2e⁻ → Sn²⁺ + 2e⁻ + 2 Ag

5) Fourth step, cancell the electrons on both sides:

Sn + 2Ag⁺ → Sn²⁺ + 2 Ag

And that is the final answer.

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l