1864 United States presidential election in Pennsylvania explained

See main article: 1864 United States presidential election.

Election Name:1864 United States presidential election in Pennsylvania
Country:Pennsylvania
Type:presidential
Ongoing:no
Previous Election:1860 United States presidential election in Pennsylvania
Previous Year:1860
Next Election:1868 United States presidential election in Pennsylvania
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 Vote1:26
Popular Vote1:296,391
Percentage1:51.75%
Nominee2:George B. McClellan
Party2:Democratic Party (United States)
Home State2:New Jersey
Running Mate2:George H. Pendleton
Electoral Vote2:0
Popular Vote2:276,316
Percentage2:48.25%
Map Size:300px
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 Pennsylvania took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose 26 representatives, or electors to the Electoral College, who voted for president and vice president.

Pennsylvania voted for the National Union candidate, incumbent Republican 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 narrow margin of 3.5%.

Results

1864 United States presidential election in Pennsylvania[1]
PartyCandidateVotesPercentageElectoral votes
National UnionAbraham Lincoln (incumbent)296,39151.75%26
DemocraticGeorge B. McClellan276,31648.25%0
Totals572,707100.0%26

See also

Notes and References

  1. Web site: 1864 Presidential General Election Results - Pennsylvania. U.S. Election Atlas. 3 August 2012.