1864 United States presidential election in Rhode Island explained

See main article: 1864 United States presidential election.

Election Name:1864 United States presidential election in Rhode Island
Country:Rhode Island
Type:presidential
Ongoing:no
Previous Election:1860 United States presidential election in Rhode Island
Previous Year:1860
Next Election:1868 United States presidential election in Rhode Island
Next Year:1868
Election Date:November 8, 1864
Image1:Abraham Lincoln November 1863.jpg
Nominee1:Abraham Lincoln
Party1:National Union Party (United States)
Home State1:Illinois
Running Mate1:Andrew Johnson
Electoral Vote1:4
Popular Vote1:13,962
Percentage1:62.24%
Nominee2:George B. McClellan
Party2:Democratic Party (United States)
Home State2:New Jersey
Running Mate2:George H. Pendleton
Electoral Vote2:0
Popular Vote2:8,470
Percentage2:37.76%
Map Size:250px
President
Before Election:Abraham Lincoln
Before Party:Republican Party (United States)
After Election:Abraham Lincoln
After Party:National Union Party (United States)

The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the National Union candidate, incumbent Republican Party President Abraham Lincoln and his running mate Andrew Johnson. They defeated the Democratic candidate, George B. McClellan and his running mate George H. Pendleton. Lincoln won the state by a margin of 24.48%.

Results

1864 United States presidential election in Rhode Island[1]
PartyCandidateVotesPercentageElectoral votes
National UnionAbraham Lincoln (incumbent)13,96262.24%4
DemocraticGeorge B. McClellan8,47037.76%0
Totals22,432100.0%4

See also

Notes and References

  1. Web site: 1864 Presidential General Election Results - Rhode Island.