What is the shorthand notation that represents the following galvanic cell reaction?

Question

What is the shorthand notation that represents the following galvanic cell reaction?

Ba(s) + 2Ag(NO₃)(aq) → Ba(NO₃)₂(aq) + 2Ag(s)

Answer 1

The question asks to write essentially the cell notation for the following reaction.

Ba(s) + 2AgNO3(aq) >>>> Ba(NO3)2(aq) + 2Ag(s)

The first issue is to determine what is oxidized (lost electrons and therefore will be more +) and what is reduced (gained electrons and will be more negative)

Looking at Ba(s) as a reactant it exists here as the free element and has no charge, As a product it has a +2 charge (Ba+2(aq) to counter 2 of the negative 1 charge on the polyatomic ion NO3-1(aq). So Ba(s) loses 2 electrons resulting in the product Ba+2(aq) therefore Ba(s) has been oxidized,

Now looking at the reactant Ag+(aq) has a + 1 charge to counter the (-) 1 charge of the NO3- ion.

Reviewing the balanced chemical reaction you need a coefficient of 2 placed in front of the reactant AgNO3 to balance the NO3- ions so a coefficient must also be placed in front of the product Ag(s) so 2 electrons has been gained by Ag+(s) and therefore Ag+(aq) has been reduced.

The half cell reactions has this appearance:

Ba(s) >>>>>>> Ba+2(a) + 2e (oxidation - loss of 2 electrons)

2Ag+1(aq) + 2e) >>>>>> 2Ag(s) (reduction - gain of 2 electrons)

In a galvanic cell, oxidation takes place at the anode where there is also a loss of mass and reduction takes place at the cathode where there is a gain of mass.

To write the cell notation what is oxidized is written first on the left side of the notation and the reduction is written on the tight side of the notation separated by two vertical lines to indicate the salt bridge where there is the passage of the ions in solution.

Ba(s) / Ba+2(aq) // Ag+(aq) / Ag(s)

The vertical line between Ba(s) and the Ba+2(aq) as well as between Ag+(aq) and Ag(s) is placed there to show there is a phase change (in this case from solid to (aq) in Barium and a phase change from (aq) to solid with regards to silver. Also note no coefficients are used in the cell notation only in the balancing of the redox reaction. Finally note there is no vertical line to the left of Ba(s) and to the right of Ag(s) because they act as the electrodes.

Answer 2

What is the shorthand notation that represents the following galvanic cell reaction?

Ba(s) + 2Ag(NO₃)(aq) → Ba(NO₃)₂(aq) + 2Ag(s)

A galvanic cell has two parts: a cathode and an anode. The anode is where oxidation occurs.

The cathode is where reduction occurs. We would have to split this reaction into two half reactions

(one that represents the oxidation that occurs at the anode, the other that represents the reduction that happens at the anode).

Step 1: Split the reaction into two half reactions.

Ba(s) → Ba(NO₃)₂(aq)

2Ag(NO₃)(aq) → 2Ag(s)

Step 2: Determine which half reaction is the oxidation reaction and which one is the reduction one. To do this, you would have to find the oxidation states for the metal atom in each half reaction.

Ba(s) → Ba(NO₃)₂(aq) Ba(s) has an oxidation state of 0 because it is a free element. Ba has an oxidation state of +2 in Ba(NO₃)₂(aq) because NO₃^- has a charge of -1 and there are two NO₃^- molecules in Ba(NO₃)₂(aq). The overall charge of Ba(NO₃)₂ is 0 so Ba must have an oxidation state of +2 to cancel out the charges from the two NO₃^-.

Because Ba goes from an oxidation state of 0 to an oxidation state of +2 in this reaction, it is oxidized (Ba loses two electrons). Ba(s) → Ba(NO₃)₂(aq) is the oxidation half reaction.

In Ag(NO₃)(aq) → 2Ag(s), silver goes from an oxidation state of +2 to an oxidation state of 0. It gains two electrons so it is reduced. This is the reduction reaction (occurs at the cathode).

Now that we have identified the cathode and anode, we can use short hand notation to represent the reactions that occur in the cell. || is used to separate the reaction that occurs in the cathode from the reaction that occurs in the anode. The correct shorthand form is shown below:

Ba(s)|Ba(NO₃)₂||AgNO₃(aq)|Ag(s)

Ba(NO₃)₂ is on the right side of the single line in the anode side (|) because it is the chemical compound in the anode that has the highest oxidation state for Ba. Ba(s) has the lowest oxidation state for Ba in the anode cell. AgNO₃(aq) is on the left side of the single line (|) on the cathode side of the shorthand notation because Ag has a higher oxidation state (+2) in Ag(NO₃)(aq) than in Ag(s) (0).