Copper and silver are in the same group (11 or copper group) and of course have some similarities.
Copper (Cu) does not form more than one type of cation.
To calculate the percent yield of silver (Ag) in the reaction, we first need to determine the theoretical yield of Ag based on the amount of Cu used. The balanced chemical equation shows that 1 mole of Cu produces 2 moles of Ag. The molar mass of Cu is approximately 63.55 g/mol, and that of Ag is about 107.87 g/mol. From 12.7 g of Cu, we can calculate the moles of Cu: [ \text{Moles of Cu} = \frac{12.7 \text{ g}}{63.55 \text{ g/mol}} \approx 0.199 \text{ moles} ] Since 1 mole of Cu produces 2 moles of Ag, the theoretical yield of Ag would be: [ 0.199 \text{ moles Cu} \times 2 \text{ moles Ag/mole Cu} \times 107.87 \text{ g/mol} \approx 43.0 \text{ g Ag} ] Now, using the actual yield of 38.1 g Ag, the percent yield can be calculated as: [ \text{Percent Yield} = \left(\frac{38.1 \text{ g}}{43.0 \text{ g}}\right) \times 100 \approx 88.8% ] Thus, the percent yield of silver in this reaction is approximately 88.8%.
The overall voltage for the redox reaction involving Ag and Cu is determined by subtracting the reduction potential of the anode from the reduction potential of the cathode. Given the reduction potentials of Ag and Cu as 0.80 V and 0.34 V respectively, the overall voltage is 0.46 V, calculated as (0.34 V) - (0.80 V).
Cu(s) + 2AgNO3(aq) ---> Cu(NO3)2(aq) + 2Ag(s) So you need half as many moles of Cu. Thus 5.8/2 = 2.9 moles of Cu are needed.
It's been a long while since chem, but perhaps this will put you on the path. When they're talking about "half reaction" I believe they're referring to one side or the other, of the balanced equation -- what forms from what. In the case of solid copper and silver nitrate, the copper "takes" the NO3 and leaves the silver, previously in solution, to remain as a solid. Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag
cu,ag, and au are all in what group
Cu + AgNO3 → Ag + Cu(NO3)2 In this reaction, copper (Cu) is more reactive than silver (Ag), so it will replace silver in the compound AgNO3, resulting in the formation of silver metal and copper nitrate.
The balanced equation for the reaction is: Cu + 2AgNO3 -> Cu(NO3)2 + 2Ag Calculate the molar mass of Cu and Ag (Cu = 63.55 g/mol, Ag = 107.87 g/mol). Using the molar ratio of Cu to Ag (1:2), convert the mass of Cu to moles, then use the molar ratio to find the moles of Ag produced. Finally, convert moles of Ag to grams using the molar mass of Ag to find the grams of silver produced.
cu(II) + 2agcl --> 2ag+cucl2
cu,ag, and au are all in what group
Copper (Cu) does not form more than one type of cation.
The correct answers are: Ag+ and Cu+2.
Group eleven on the periodic table consists of the transition metals copper (Cu), silver (Ag), and gold (Au). These elements share similar properties such as high thermal and electrical conductivity, luster, and resistance to corrosion.
To calculate the percent yield of silver (Ag) in the reaction, we first need to determine the theoretical yield of Ag based on the amount of Cu used. The balanced chemical equation shows that 1 mole of Cu produces 2 moles of Ag. The molar mass of Cu is approximately 63.55 g/mol, and that of Ag is about 107.87 g/mol. From 12.7 g of Cu, we can calculate the moles of Cu: [ \text{Moles of Cu} = \frac{12.7 \text{ g}}{63.55 \text{ g/mol}} \approx 0.199 \text{ moles} ] Since 1 mole of Cu produces 2 moles of Ag, the theoretical yield of Ag would be: [ 0.199 \text{ moles Cu} \times 2 \text{ moles Ag/mole Cu} \times 107.87 \text{ g/mol} \approx 43.0 \text{ g Ag} ] Now, using the actual yield of 38.1 g Ag, the percent yield can be calculated as: [ \text{Percent Yield} = \left(\frac{38.1 \text{ g}}{43.0 \text{ g}}\right) \times 100 \approx 88.8% ] Thus, the percent yield of silver in this reaction is approximately 88.8%.
Copper (Cu), silver (Ag), and gold (Au) are transition metals that have high electrical conductivity and are commonly used in jewelry and electrical wiring. Aluminum (Al) is a lightweight metal commonly used in construction due to its strength and resistance to corrosion.
Because the valence of both Cu and Ni is +2, but Ag have a valence of +1 and the metals with lower valence do not form solid solution when added to the metals with higher valence.
The requested equation is Cu + AgNO3 -> CuNO3 + Ag or Cu + 2 AgNO3 -> Cu(NO3)2 + 2 Ag, depending on whether copper forms its (II) or (I) cations.