looks like a single displacement reaction. Cu + AgSO4 Thank you very much!
This is a displacement reaction: Copper is (barely) higher in the electromotive series than silver, so that the silver in a compound can be displaced by copper, resulting in the formation of unreacted silver.
0.80-0.34
It would produce Silver and a blue solid called Copper Nitrate (Cu(NO3)2). It is caused by a replacement reaction where Copper replaced Silver in Copper Nitrate. The equation goes like this: Cu + Ag(NO3)2 → Ag + Cu(NO3)2
In the reaction Zn + CuCl2 → ZnCl2 + Cu, CuCl2 is the oxidizing agent because it accepts electrons from Zn, causing zinc to be oxidized and copper to be reduced.CuCl2 itself gets reduced to Cu.
Cl2 + 2Cu --> 2CuCl Oxidation reaction is Cu --> Cu+ + 1e Reduction reaction is Cl + 1e --> Cl- Redox reaction is Cu + Cl --> Cu+ + Cl-
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 reaction is:Cu + 2 AgNO3 = Cu(NO3)2 + 2 Ag
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.
In the cell, the half-reaction for silver will be Ag+ (aq) + e- -> Ag (s) with a standard reduction potential of +0.80 V. The half-reaction for copper will be Cu2+ (aq) + 2e- -> Cu (s) with a standard reduction potential of +0.34 V. The silver half-reaction will occur at the cathode, while the copper half-reaction will occur at the anode in the cell.
This reaction involves the reaction of copper (Cu) with silver nitrate (AgNO3) to form copper(II) nitrate (Cu(NO3)2) and silver (Ag). It is a chemical reaction that was likely conducted in a laboratory setting or for academic purposes.
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This is a displacement reaction: Copper is (barely) higher in the electromotive series than silver, so that the silver in a compound can be displaced by copper, resulting in the formation of unreacted silver.
cu(II) + 2agcl --> 2ag+cucl2
0.80-0.34
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.
To determine the overall voltage for the redox reaction involving the half-reactions ( \text{Ag}^+ + e^- \rightarrow \text{Ag}(s) ) and ( \text{Cu}(s) \rightarrow \text{Cu}^{2+} + 2e^- ), we first need the standard reduction potentials. The standard reduction potential for silver (( \text{Ag}^+ )) is +0.80 V, and for copper (( \text{Cu}^{2+} )) is +0.34 V. Since silver is reduced and copper is oxidized, the overall cell potential is calculated as ( E_{\text{cell}} = E_{\text{reduction}} - E_{\text{oxidation}} = 0.80 , \text{V} - 0.34 , \text{V} = 0.46 , \text{V} ). Thus, the overall voltage for the redox reaction is +0.46 V.
There is NO reaction between Cu and Ag.