Ba(s) has the lowest oxidation state for Ba in the anode cell. The correct shorthand form is shown below:īa(NO 3) 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. || is used to separate the reaction that occurs in the cathode from the reaction that occurs in the anode. Now that we have identified the cathode and anode, we can use short hand notation to represent the reactions that occur in the cell. This is the reduction reaction (occurs at the cathode). In Ag(NO 3)(aq) → 2Ag(s), silver goes from an oxidation state of +2 to an oxidation state of 0. Ba(s) → Ba(NO 3) 2(aq) is the oxidation half reaction. The overall charge of Ba(NO 3) 2 is 0 so Ba must have an oxidation state of +2 to cancel out the charges from the two NO 3 -.īecause 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 has an oxidation state of +2 in Ba(NO 3) 2(aq) because NO 3 - has a charge of -1 and there are two NO 3 - molecules in Ba(NO 3) 2(aq). To do this, you would have to find the oxidation states for the metal atom in each half reaction.īa(s) → Ba(NO 3) 2(aq) Ba(s) has an oxidation state of 0 because it is a free element. Step 2: Determine which half reaction is the oxidation reaction and which one is the reduction one. Step 1: Split the 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). We would have to split this reaction into two half reactions What is the shorthand notation that represents the following galvanic cell reaction?īa(s) + 2Ag(NO 3)(aq) → Ba(NO 3) 2(aq) + 2Ag(s)Ī galvanic cell has two parts: a cathode and an anode. 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. Also note no coefficients are used in the cell notation only in the balancing of the redox reaction. 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. 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. 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. The half cell reactions has this appearance:īa(s) > Ba+2(a) + 2e (oxidation - loss of 2 electrons)ĢAg+1(aq) + 2e) > 2Ag(s) (reduction - gain of 2 electrons) 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. Now looking at the reactant Ag+(aq) has a + 1 charge to counter the (-) 1 charge of the NO3- ion. So Ba(s) loses 2 electrons resulting in the product Ba+2(aq) therefore Ba(s) has been oxidized, 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). 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) One of the simplest cells is the Daniell cell.The question asks to write essentially the cell notation for the following reaction.īa(s) + 2AgNO3(aq) > Ba(NO3)2(aq) + 2Ag(s) When known, the initial concentrations of the various ions are usually included. Note that spectator ions are not included and that the simplest form of each half-reaction was used.
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