1864 United States presidential election in Rhode Island

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1864 United States presidential election in Rhode Island

← 1860 November 8, 1864 1868 →
  Abraham Lincoln November 1863.jpg GeorgeMcClellan2.jpg
Nominee Abraham Lincoln George B. McClellan
Party National Union Democratic
Home state Illinois New Jersey
Running mate Andrew Johnson George H. Pendleton
Electoral vote 4 0
Popular vote 13,962 8,470
Percentage 62.24% 37.76%

President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

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, Abraham Lincoln, over the Democratic candidate, George B. McClellan. Lincoln won the state by a margin of 24.48%.

Results[edit]

1864 United States presidential election in Rhode Island[1]
Party Candidate Votes Percentage Electoral votes
National Union Abraham Lincoln 13,962 62.24% 4
Democratic George B. McClellan 8,470 37.76% 0
Totals 22,432 100.0% 4

References[edit]

  1. ^ "1864 Presidential General Election Results - Rhode Island".